Ellipse for JEE: Notes, Eccentric Angle, Tangents, Normals and PYQs

An ellipse is the path of a planet round the sun, the outline of a circle seen at an angle and the shape of a whispering gallery. In coordinate geometry it is the closed conic: two foci, two directrices, eccentricity between $0$ and $1$. Most JEE questions on it reduce to one idea, the eccentric angle point $(a\cos\theta,b\sin\theta)$, which turns chords, tangents and normals into short relations between angles.

This note covers the two definitions of the ellipse, the standard ellipse and its parts, both orientations, the auxiliary circle and the eccentric angle, shifted and slanted ellipses, the position of a point, a line meeting the ellipse, chords, focal chords, chords with a given middle point, diameters and conjugate diameters, tangents in all three forms, pairs of tangents, the chord of contact, the director circle, pole and polar, common tangents, normals and co-normal points, the reflection property, the area of an ellipse and locus problems. The worked examples include JEE Main 2025 and 2026 questions, earlier JEE Main questions, and JEE Advanced and IIT JEE questions; every answer was re-derived from scratch.

Conic sections and the ellipse

Let $S$ be a fixed point and $\ell$ a fixed line not through $S$. The locus of a point $P$ that moves so that

$$\frac{PS}{PM}=e$$

where $PM$ is the perpendicular distance of $P$ from $\ell$, is a conic section with focus $S$, directrix $\ell$ and eccentricity $e$.

So an ellipse is the locus of a point whose distance from a fixed point is a fixed fraction $e<1$ of its distance from a fixed line:

$$PS=e\cdot PM,\qquad 0<e<1$$ An ellipse with its focus S on the major axis, the directrix as a vertical line on the right, a point P on the curve joined to S and to the foot M of its perpendicular on the directrix

The second definition: sum of focal distances

An ellipse has two foci $S$ and $S'$. For every point $P$ on it,

$$PS+PS'=2a$$

where $2a$ is the length of the major axis. Conversely, the locus of a point whose distances from two fixed points $S$, $S'$ add up to a constant $2a>SS'$ is an ellipse with foci $S$ and $S'$.

An ellipse with both foci S and S' marked on the major axis and a point P on the curve joined to both foci, the two focal distances drawn as segments

This second form is the fastest way to spot an ellipse in a locus question or in a complex number condition $|z-z_1|+|z-z_2|=k$.

Example 1 (IITianForum). Find the equation of the ellipse whose foci are $(\pm4,0)$ and for which the sum of the focal distances of every point is $10$.

▸ SHOW SOLUTION

Step 1. Write the condition

For $P(x,y)$:

$$\sqrt{(x-4)^2+y^2}=10-\sqrt{(x+4)^2+y^2}$$

Step 2. Square once

The $x^2$ and $y^2$ terms cancel:

$$-16x=100-20\sqrt{(x+4)^2+y^2}$$$$5\sqrt{(x+4)^2+y^2}=25+4x$$

Step 3. Square again

$$25\left(x^2+8x+16+y^2\right)$$$$=625+200x+16x^2$$$$9x^2+25y^2=225$$

Answer: $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$

Example 2 (JEE Main 2026, 2 April Shift 2). Let $A$ be the point $(3,0)$ and circles with variable diameter $AB$ touch the circle $x^2+y^2=36$ internally. Let the curve $C$ be the locus of the point $B$. If the eccentricity of $C$ is $e$, then $72e^2$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Circle on AB

Let $B=(h,k)$. Centre $M=\left(\frac{h+3}2,\frac k2\right)$, radius $r=\frac{AB}2$.

Step 2. Internal contact

$OM=6-r$. Multiply by $2$; $2\,OM$ is the distance of $B$ from $A'(-3,0)$:

$$BA'=12-BA$$$$BA+BA'=12$$

Step 3. Read the ellipse

Foci $(\pm3,0)$ and $2a=12$, so $a=6$, $ae=3$, $e=\frac12$.

$$72e^2=72\cdot\frac14=18$$

Answer: $18$

Example 3 (JEE Main 2026, 8 April Shift 2). The number of values of $z\in\mathbb{C}$ satisfying $|z-(4+8i)|=\sqrt{10}$ and $|z-(3+5i)|+|z-(5+11i)|=4\sqrt5$ is

▸ SHOW SOLUTION

Step 1. The ellipse

Foci $(3,5)$ and $(5,11)$, $2a=4\sqrt5$, so $a^2=20$. Centre $(4,8)$, half focal distance

$$ae=\frac12\sqrt{4+36}=\sqrt{10}$$$$b^2=a^2-a^2e^2=20-10=10$$

Step 2. The circle

Centre $(4,8)$, the centre of the ellipse, radius $\sqrt{10}=b$.

Step 3. Count

Every point of the ellipse is at distance at least $b$ from the centre, with equality only at the two ends of the minor axis. So the circle meets the ellipse at exactly these $2$ points.

Answer: option (2)

Example 4 (IIT JEE 2009, Paper 1). In a triangle $ABC$ with fixed base $BC$, the vertex $A$ moves such that $\cos B+\cos C=4\sin^2\frac A2$. If $a,b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A,B$ and $C$ respectively, then

▸ SHOW SOLUTION

Step 1. Simplify the condition

$$2\cos\frac{B+C}2\cos\frac{B-C}2=4\sin^2\frac A2$$

Since $\cos\frac{B+C}2=\sin\frac A2\ne0$:

$$\cos\frac{B-C}2=2\sin\frac A2$$

Step 2. Use the sine rule

$$\frac{b+c}a=\frac{\sin B+\sin C}{\sin A}=\frac{\cos\frac{B-C}2}{\sin\frac A2}=2$$

So $b+c=2a$.

Step 3. Locus

$AC+AB=2\,BC$, a constant larger than $BC$. So $A$ moves on an ellipse with foci $B$ and $C$ (the two points on line $BC$ excluded).

Answer: (B) and (C)

Example 5 (IITianForum). The values of $a$ for which $(5x-1)^2+(5y-2)^2=\left(a^2-3\right)(3x+4y-1)^2$ represents an ellipse are

▸ SHOW SOLUTION

Step 1. Put it in PS = e PM form

Divide by $25$:

$$PS^2=\left(x-\frac15\right)^2+\left(y-\frac25\right)^2$$$$PS^2=\left(a^2-3\right)\left(\frac{3x+4y-1}5\right)^2$$

Here $S=\left(\frac15,\frac25\right)$, and the bracket is $PM$, the distance of $P$ from $3x+4y-1=0$.

Step 2. Check the focus is off the line

$3\cdot\frac15+4\cdot\frac25-1=\frac65\ne0$. So $e^2=a^2-3$.

Step 3. Ellipse condition

$$0<a^2-3<1\iff3<a^2<4$$

So $a\in\left(-2,-\sqrt3\right)\cup\left(\sqrt3,2\right)$.

Answer: (A) and (B)

Example 6 (JEE Main 2026, 6 April Shift 1). Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2-ax+2=0$. Let the set $\{a\in\mathbb{R}:e_1,e_2$ are eccentricities of hyperbolas$\}=(\alpha,\beta)$ and $\{a\in\mathbb{R}:e_1,e_2$ are the eccentricities of an ellipse and a hyperbola respectively$\}=(\gamma,\infty)$. Then $\alpha^2+\beta^2+\gamma^2$ is equal to

▸ SHOW SOLUTION

Step 1. Both roots greater than 1

Let $f(x)=x^2-ax+2$. Distinct real roots: $a^2>8$. Both roots exceed $1$: $f(1)=3-a>0$ and $\frac a2>1$. So

$$2\sqrt2<a<3$$

Step 2. One root in (0, 1), the other above 1

$f(0)=2>0$ and $f(1)=3-a<0$, so $a>3$. Hence $\gamma=3$.

Step 3. Add

$$\alpha^2+\beta^2+\gamma^2=8+9+9=26$$

Answer: option (3)

The standard ellipse

Take the centre at the origin, the focus $S(ae,0)$ and the directrix $x=\frac ae$, where $a>0$ and $0<e<1$. For $P(x,y)$, $PS=e\cdot PM$ gives

$$(x-ae)^2+y^2=e^2\left(\frac ae-x\right)^2$$ $$x^2\left(1-e^2\right)+y^2=a^2\left(1-e^2\right)$$

Put $b^2=a^2\left(1-e^2\right)$:

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\qquad a>b>0$$

This is the standard ellipse. Since the equation is unchanged when $x\to-x$ and when $y\to-y$, the curve is symmetric about both axes, so there is a second focus $S'(-ae,0)$ with a second directrix $x=-\frac ae$.

$$e^2=1-\frac{b^2}{a^2}$$

Parts of the ellipse

The standard ellipse with centre C, vertices A and A', ends of the minor axis B and B', foci S and S', both directrices dashed, and the latus rectum LL' through S

• Centre $C(0,0)$: every chord through it is bisected there.
• Major axis $AA'$: from $A'(-a,0)$ to $A(a,0)$, length $2a$.
• Minor axis $BB'$: from $B'(0,-b)$ to $B(0,b)$, length $2b$.
• Vertices: $A$ and $A'$, the ends of the major axis.
• Foci: $S(ae,0)$ and $S'(-ae,0)$; distance between foci $2ae$.
• Directrices: $x=\pm\frac ae$, at distance $\frac{2a}e$ apart.
• Latus rectum: the focal chord perpendicular to the major axis.

Since $b^2=a^2-a^2e^2$, the three lengths $a$, $b$, $ae$ form a right triangle: the distance from an end of the minor axis to a focus is $a$.

$$BS=\sqrt{b^2+a^2e^2}=a$$

Latus rectum and focal distances

Put $x=ae$ in the equation: $y^2=b^2\left(1-e^2\right)=\frac{b^4}{a^2}$. The ends of the latus rectum are

$$L\left(ae,\frac{b^2}a\right),\qquad L'\left(ae,-\frac{b^2}a\right)$$

and its length is $\frac{2b^2}a$. The half length $\frac{b^2}a=a\left(1-e^2\right)$ is the semi-latus rectum.

For $P(x_1,y_1)$ on the ellipse, $PS=e\cdot PM$ with $PM=\frac ae-x_1$, and similarly for $S'$:

$$SP=a-ex_1,\qquad S'P=a+ex_1$$

Adding gives $SP+S'P=2a$, the second definition. Multiplying gives a result used often:

$$SP\cdot S'P=a^2-e^2x_1^2$$

Two orientations

If $b>a$ in $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, the major axis is along the $y$-axis. Every result is obtained by interchanging the roles of $x$ and $y$, and of $a$ and $b$.

Two ellipses side by side: the first wider than tall with foci on the x-axis, the second taller than wide with foci on the y-axis, each with its directrices dashed

Example 7 (IITianForum). For the ellipse $9x^2+25y^2=225$ find the eccentricity, the foci, the directrices, the length of the latus rectum and the focal distances of the point $\left(4,\frac95\right)$.

▸ SHOW SOLUTION

Step 1. Standard form

$$\frac{x^2}{25}+\frac{y^2}9=1$$

$a=5$, $b=3$, $a>b$.

Step 2. The parts

$$e^2=1-\frac9{25}=\frac{16}{25},\qquad e=\frac45$$

Foci $(\pm4,0)$, directrices $x=\pm\frac{25}4$, latus rectum $\frac{2\cdot9}5=\frac{18}5$.

Step 3. Focal distances

$$SP=5-\frac45\cdot4=\frac95$$$$S'P=5+\frac{16}5=\frac{41}5$$

(The point is an end of the latus rectum, so $SP$ is the semi-latus rectum.)

Answer: $e=\frac45$, foci $(\pm4,0)$, $x=\pm\frac{25}4$, latus rectum $\frac{18}5$, focal distances $\frac95$ and $\frac{41}5$

Example 8 (JEE Main 2026, 5 April Shift 1). Let a focus of the ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ be $S(4,0)$ and its eccentricity be $\frac45$. If the point $P(3,\alpha)$ lies on $E$ and $O$ is the origin, then the area of $\triangle POS$ is

▸ SHOW SOLUTION

Step 1. The ellipse

$ae=4$, $e=\frac45$, so $a=5$ and $b^2=25-16=9$.

Step 2. The point P

$$\frac9{25}+\frac{\alpha^2}9=1,\qquad|\alpha|=\frac{12}5$$

Step 3. Area

Base $OS=4$, height $|\alpha|$:

$$\frac12\cdot4\cdot\frac{12}5=\frac{24}5$$

Answer: option (3)

Example 9 (JEE Main 2026, 24 January Shift 2). Let the length of the latus rectum of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $(a>b)$, be $30$. If its eccentricity equals the maximum value of the function $f(t)=-\frac34+2t-t^2$, then $\left(a^2+b^2\right)$ is equal to

▸ SHOW SOLUTION

Step 1. Eccentricity

$$f(t)=\frac14-(t-1)^2\le\frac14$$

So $e=\frac14$ and $b^2=a^2\left(1-\frac1{16}\right)=\frac{15a^2}{16}$.

Step 2. Latus rectum

$$\frac{2b^2}a=\frac{15a}8=30,\qquad a=16$$

$b^2=240$.

Step 3. Add

$$a^2+b^2=256+240=496$$

Answer: option (3)

Example 10 (JEE Main 2026, 22 January Shift 2). Let $S$ and $S'$ be the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ and $P(\alpha,\beta)$ be a point on the ellipse in the first quadrant. If $(SP)^2+(S'P)^2-SP\cdot S'P=37$, then $\alpha^2+\beta^2$ is equal to

▸ SHOW SOLUTION

Step 1. Focal distances

$e=\frac45$, so $SP=5-\frac{4\alpha}5$ and $S'P=5+\frac{4\alpha}5$.

Step 2. Use sum and product

$SP+S'P=10$ and $SP\cdot S'P=25-\frac{16\alpha^2}{25}$:

$$(SP+S'P)^2-3\,SP\cdot S'P=37$$$$100-75+\frac{48\alpha^2}{25}=37,\qquad\alpha^2=\frac{25}4$$

Step 3. The ordinate

$$\beta^2=9\left(1-\frac14\right)=\frac{27}4$$$$\alpha^2+\beta^2=\frac{52}4=13$$

Answer: option (4)

Example 11 (JEE Main 2025, 2 April Shift 1). If $S$ and $S'$ are the foci of the ellipse $\frac{x^2}{18}+\frac{y^2}{9}=1$ and $P$ is a point on the ellipse, then $\min(SP\cdot S'P)+\max(SP\cdot S'P)$ is equal to

▸ SHOW SOLUTION

Step 1. The product

$e^2=1-\frac9{18}=\frac12$. For $P(x,y)$:

$$SP\cdot S'P=a^2-e^2x^2=18-\frac{x^2}2$$

Step 2. Range

$0\le x^2\le18$, so the product runs from $9$ (at the vertices) to $18$ (at the ends of the minor axis).

$$9+18=27$$

Answer: option (4)

Example 12 (IITianForum). The length of the major axis of an ellipse is $40$ and its eccentricity is $\frac12$. The four end points of its two latera recta are joined to make a rectangle. The area of this rectangle is

▸ SHOW SOLUTION

Step 1. The lengths

$a=20$, $ae=10$, $b^2=400\left(1-\frac14\right)=300$.

Step 2. Sides of the rectangle

One side is the distance between the foci, $2ae=20$. The other is the latus rectum:

$$\frac{2b^2}a=\frac{600}{20}=30$$

Step 3. Area

$$20\times30=600$$

Answer: option (3)

Example 13 (JEE Main 2026, 28 January Shift 2). Let $A=\left\{z\in\mathbb{C}:|z-2|\le4\right\}$ and $B=\left\{z\in\mathbb{C}:|z-2|+|z+2|=5\right\}$. Then the maximum of $\left\{|z_1-z_2|:z_1\in A,\ z_2\in B\right\}$ is

▸ SHOW SOLUTION

Step 1. Read the sets

$A$ is the disc of radius $4$ about $2$. $B$ is the ellipse with foci $\pm2$ and $2a=5$, so $a=\frac52$.

Step 2. Largest distance from a point of the disc

For a fixed $z_2$, the farthest point of $A$ is at distance $|z_2-2|+4$.

Step 3. Largest focal distance

$|z_2-2|$ is the distance of $z_2$ from the focus $2$. Its largest value is $a+ae=\frac52+2=\frac92$, at the far vertex $-\frac52$.

$$\frac92+4=\frac{17}2$$

Answer: option (3)

Example 14 (JEE Main 2026, 2 April Shift 1). Let an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a<b$, pass through the point $(4,3)$ and have eccentricity $\frac{\sqrt5}3$. Then the length of its latus rectum is

▸ SHOW SOLUTION

Step 1. Use the a < b formula

$$\frac59=1-\frac{a^2}{b^2},\qquad a^2=\frac49b^2$$

Step 2. Through (4, 3)

$$\frac{36}{b^2}+\frac9{b^2}=1,\qquad b^2=45,\ a^2=20$$

Step 3. Latus rectum

The major axis is vertical, so the latus rectum is $\frac{2a^2}b$:

$$\frac{40}{3\sqrt5}=\frac{8\sqrt5}3$$

Answer: option (4)

Example 15 (JEE Main 2026, 24 January Shift 1). Let each of the two ellipses $E_1:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ and $E_2:\frac{x^2}{A^2}+\frac{y^2}{B^2}=1$ $(A<B)$ have eccentricity $\frac45$. Let the lengths of the latera recta of $E_1$ and $E_2$ be $\ell_1$ and $\ell_2$ respectively, such that $2\ell_1^2=9\ell_2$. If the distance between the foci of $E_1$ is $8$, then the distance between the foci of $E_2$ is ______.

▸ SHOW SOLUTION

Step 1. The first ellipse

$2ae=8$, $e=\frac45$: $a=5$, $b^2=9$, so $\ell_1=\frac{18}5$.

Step 2. The second ellipse

$A^2=B^2\left(1-\frac{16}{25}\right)=\frac{9B^2}{25}$, and $\ell_2=\frac{2A^2}B=\frac{18B}{25}$.

Step 3. Use the relation

$$2\cdot\frac{324}{25}=9\cdot\frac{18B}{25},\qquad B=4$$

Distance between the foci of $E_2$:

$$2Be=2\cdot4\cdot\frac45=\frac{32}5$$

Answer: $\frac{32}5$

Example 16 (JEE Main 2026, 8 April Shift 2). Let $\frac{x^2}{f\left(a^2+7a+3\right)}+\frac{y^2}{f(3a+15)}=1$ represent an ellipse with major axis along the $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbb{R}$. If the set of all possible values of $a$ is $\mathbb{R}-[\alpha,\beta]$, then $\alpha^2+\beta^2$ equals

▸ SHOW SOLUTION

Step 1. Major axis along the y-axis

The $y^2$ denominator must be larger:

$$f(3a+15)>f\left(a^2+7a+3\right)$$

Step 2. Remove f

$f$ is strictly decreasing, so the inequality reverses:

$$a^2+7a+3>3a+15$$$$(a+6)(a-2)>0$$

Step 3. Read the set

$a\in\mathbb{R}-[-6,2]$, so $\alpha^2+\beta^2=36+4=40$.

Answer: option (2)

Example 17 (JEE Main 2026, 4 April Shift 1). Consider the parabola $P:y^2=4kx$ and the ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Let the line segment joining the points of intersection of $P$ and $E$ be their latus rectums. If the eccentricity of $E$ is $e$, then $e^2+2\sqrt2$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Same focus

The common latus rectum passes through both foci: $k=ae$.

Step 2. Same length

$$4k=\frac{2b^2}a\ \Rightarrow\ 4a^2e=2a^2\left(1-e^2\right)$$$$e^2+2e-1=0,\qquad e=\sqrt2-1$$

Step 3. Evaluate

$$e^2+2\sqrt2=3-2\sqrt2+2\sqrt2=3$$

Answer: $3$

Example 18 (IIT JEE 2012, Paper 1). The ellipse $E_1:\frac{x^2}9+\frac{y^2}4=1$ is inscribed in a rectangle $R$ whose sides are parallel to the coordinate axes. Another ellipse $E_2$ passing through the point $(0,4)$ circumscribes the rectangle $R$. The eccentricity of the ellipse $E_2$ is

▸ SHOW SOLUTION

Step 1. The rectangle

Its sides are the tangents $x=\pm3$, $y=\pm2$; its vertices are $(\pm3,\pm2)$.

Step 2. The second ellipse

Take $E_2:\frac{x^2}{p^2}+\frac{y^2}{16}=1$ (through $(0,4)$). Through $(3,2)$:

$$\frac9{p^2}+\frac14=1,\qquad p^2=12$$

Step 3. Eccentricity

$16>12$, so the major axis is vertical:

$$e^2=1-\frac{12}{16}=\frac14,\qquad e=\frac12$$

Answer: option (C)

Auxiliary circle and eccentric angle

The circle on the major axis as diameter,

$$x^2+y^2=a^2$$

is the auxiliary circle. Take $Q(a\cos\theta,a\sin\theta)$ on it and drop the perpendicular $QN$ to the major axis. It meets the ellipse at

$$P(a\cos\theta,\ b\sin\theta)$$

The angle $\theta$ is the eccentric angle of $P$, and $P$ is called the point $\theta$.

An ellipse inside its auxiliary circle, a point Q on the circle joined to the centre at angle theta, the perpendicular from Q to the major axis cutting the ellipse at P

Since $\frac{PN}{QN}=\frac ba$, the ellipse is the auxiliary circle compressed towards the major axis in the ratio $b:a$. Every vertical chord of the circle is cut by the ellipse in this ratio.

Parametric form

For every $\theta\in[0,2\pi)$ the point

$$x=a\cos\theta,\qquad y=b\sin\theta$$

lies on the ellipse, and every point arises from exactly one $\theta$. In this form

$$SP=a(1-e\cos\theta)$$ $$S'P=a(1+e\cos\theta)$$

Example 19 (IIT JEE 2009, Paper 1). The line passing through the extremity $A$ of the major axis and extremity $B$ of the minor axis of the ellipse $x^2+9y^2=9$ meets its auxiliary circle at the point $M$. Then the area of the triangle with vertices at $A$, $M$ and the origin $O$ is

▸ SHOW SOLUTION

Step 1. The points

$\frac{x^2}9+y^2=1$: $A(3,0)$, $B(0,1)$, auxiliary circle $x^2+y^2=9$. Line $AB$: $x=3-3y$.

Step 2. Second meeting with the circle

$$(3-3y)^2+y^2=9\ \Rightarrow\ 10y^2-18y=0$$

$y=\frac95$, so $M\left(-\frac{12}5,\frac95\right)$.

Step 3. Area

Base $OA=3$, height $\frac95$:

$$\frac12\cdot3\cdot\frac95=\frac{27}{10}$$

Answer: option (D)

Example 20 (JEE Advanced 2025, Paper 2). Let $P(x_1,y_1)$ and $Q(x_2,y_2)$ be two distinct points on the ellipse $\frac{x^2}9+\frac{y^2}4=1$ such that $y_1>0$, and $y_2>0$. Let $C$ denote the circle $x^2+y^2=9$, and $M$ be the point $(3,0)$. Suppose the line $x=x_1$ intersects $C$ at $R$, and the line $x=x_2$ intersects $C$ at $S$, such that the $y$-coordinates of $R$ and $S$ are positive. Let $\angle ROM=\frac\pi6$ and $\angle SOM=\frac\pi3$, where $O$ denotes the origin $(0,0)$. Let $|XY|$ denote the length of the line segment $XY$. Then which of the following statements is (are) TRUE?

▸ SHOW SOLUTION

Step 1. Recognise the eccentric angles

$C$ is the auxiliary circle, so $R$, $S$ give the eccentric angles of $P$, $Q$:

$$P\left(\frac{3\sqrt3}2,1\right),\qquad Q\left(\frac32,\sqrt3\right)$$

Step 2. Line PQ

At $P$: $2\cdot\frac{3\sqrt3}2+3\cdot1=3\sqrt3+3$. At $Q$: $3+3\sqrt3$. So (A) holds and (B) fails ($2x+y$ at $P$ is $3\sqrt3+1$).

Step 3. Ratios on the ordinates

Every ordinate of the ellipse is $\frac ba=\frac23$ of the circle's ordinate:

$$|N_2Q|=\frac23|N_2S|,\qquad|N_1P|=\frac23|N_1R|$$

So (C) holds; (D) would need the ratio $\frac49$, so it fails.

Answer: (A) and (C)

Example 21 (JEE Main 2026, 24 January Shift 2). Let $(h,k)$ lie on the circle $C:x^2+y^2=4$ and the point $(2h+1,3k+2)$ lie on an ellipse with eccentricity $e$. Then the value of $\frac5{e^2}$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Parametrise the circle

$h=2\cos\theta$, $k=2\sin\theta$. The point is

$$x=1+4\cos\theta,\qquad y=2+6\sin\theta$$

Step 2. The ellipse

$$\frac{(x-1)^2}{16}+\frac{(y-2)^2}{36}=1$$

Step 3. Eccentricity

The major axis is vertical:

$$e^2=1-\frac{16}{36}=\frac59,\qquad\frac5{e^2}=9$$

Answer: $9$

Example 22 (JEE Main 2026, 4 April Shift 2). Let $P(3\cos\alpha,2\sin\alpha)$, $\alpha\ne0$, be a point on the ellipse $\frac{x^2}9+\frac{y^2}4=1$, $Q$ a point on the circle $x^2+y^2-14x-14y+82=0$, and $R$ a point on the line $x+y=5$, such that the centroid of $\triangle PQR$ is $\left(2+\cos\alpha,\ 3+\frac23\sin\alpha\right)$. Then the sum of the ordinates of all possible points $R$ is

▸ SHOW SOLUTION

Step 1. Points

The circle has centre $(7,7)$, radius $4$: $Q(7+4\cos\theta,7+4\sin\theta)$. Take $R(5-y,y)$.

Step 2. Centroid

The terms in $\alpha$ cancel:

$$4\cos\theta=y-6,\qquad4\sin\theta=2-y$$

Step 3. Eliminate

$$(y-6)^2+(y-2)^2=16$$$$y^2-8y+12=0$$

$y=2$ or $6$; the sum is $8$.

Answer: option (4)

Example 23 (JEE Main 2020, 5 September Shift 1). If the point $P$ on the curve $4x^2+5y^2=20$ is farthest from the point $Q(0,-4)$, then $PQ^2$ is equal to

▸ SHOW SOLUTION

Step 1. Parametrise

$\frac{x^2}5+\frac{y^2}4=1$: $P\left(\sqrt5\cos\theta,2\sin\theta\right)$.

Step 2. Distance squared

$$PQ^2=5\cos^2\theta+(2\sin\theta+4)^2$$$$=21+16\sin\theta-\sin^2\theta$$

Step 3. Maximise

With $s=\sin\theta\in[-1,1]$, $21+16s-s^2$ increases on $[-1,1]$, so the maximum is at $s=1$:

$$21+16-1=36$$

Answer: option (4)

General forms of the ellipse

Axes parallel to the coordinate axes

Shift the centre to $(h,k)$:

$$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$$

Put $X=x-h$, $Y=y-k$; then every result of the standard ellipse holds in $X$, $Y$. For $a>b$:

An equation $Ax^2+By^2+2Gx+2Fy+C=0$ with $A$, $B$ of the same sign is turned into this form by completing the square in $x$ and in $y$.

A shifted ellipse with centre C(2, -1), the shifted axes dashed through C, both foci and both directrices marked

Ellipse from a focus and a directrix

Given a focus $S(\alpha,\beta)$, a directrix $lx+my+n=0$ and $e$, the ellipse is

$$(x-\alpha)^2+(y-\beta)^2=e^2\frac{(lx+my+n)^2}{l^2+m^2}$$

Ellipse with slanted axes

If the major axis lies along $L_1=0$ and the minor axis along $L_2=0$ (two perpendicular lines, each written with its normal of unit length), then

$$\frac{L_2^2}{a^2}+\frac{L_1^2}{b^2}=1$$

because the distance from the minor axis plays the part of $x$ and the distance from the major axis plays the part of $y$.

Recognising an ellipse

The general second degree equation

$$ax^2+2hxy+by^2+2gx+2fy+c=0$$

is an ellipse when $h^2<ab$ and the equation does not split into a pair of lines:

$$abc+2fgh-af^2-bg^2-ch^2\ne0$$

(A real ellipse also needs at least one real point; for example $x^2+y^2+1=0$ passes both tests and has none.)

Example 24 (JEE Main 2026, 28 January Shift 2). An ellipse has its centre at $(1,-2)$, one focus at $(3,-2)$ and one vertex at $(5,-2)$. Then the length of its latus rectum is

▸ SHOW SOLUTION

Step 1. Read a and ae

The major axis is the line $y=-2$. Centre to focus: $ae=2$. Centre to vertex: $a=4$.

Step 2. b squared

$$b^2=a^2-a^2e^2=16-4=12$$

Step 3. Latus rectum

$$\frac{2b^2}a=\frac{24}4=6$$

Answer: option (2)

Example 25 (JEE Main 2025, 4 April Shift 1). The length of the latus rectum of the ellipse, whose foci are $(2,5)$ and $(2,-3)$ and eccentricity is $\frac45$, is

▸ SHOW SOLUTION

Step 1. Orientation

The foci lie on $x=2$, so the major axis is vertical. Centre $(2,1)$, half focal distance $4$.

Step 2. Semi-axes

Major semi-axis $p$: $pe=4$, so $p=5$. Minor semi-axis $q$: $q^2=25-16=9$.

Step 3. Latus rectum

$$\frac{2q^2}p=\frac{18}5$$

Answer: option (4)

Example 26 (IITianForum). For the ellipse $4x^2+9y^2-16x+18y-11=0$, find the centre, the eccentricity, the foci, the directrices and the length of the latus rectum.

▸ SHOW SOLUTION

Step 1. Complete the squares

$$4\left(x^2-4x\right)+9\left(y^2+2y\right)=11$$$$4(x-2)^2+9(y+1)^2=36$$$$\frac{(x-2)^2}9+\frac{(y+1)^2}4=1$$

Step 2. Lengths

$a=3$, $b=2$:

$$e^2=1-\frac49=\frac59$$$$ae=\sqrt5,\qquad\frac ae=\frac9{\sqrt5}$$

Step 3. Points and lines

Centre $(2,-1)$, foci $\left(2\pm\sqrt5,-1\right)$, directrices $x=2\pm\frac9{\sqrt5}$, latus rectum $\frac{2\cdot4}3=\frac83$.

Answer: centre $(2,-1)$, $e=\frac{\sqrt5}3$, foci $\left(2\pm\sqrt5,-1\right)$, directrices $x=2\pm\frac9{\sqrt5}$, latus rectum $\frac83$

A slanted ellipse with its major axis along the line y equals x, the two vertex tangents perpendicular to that axis, a circle whose diameter MN lies on a line touching the ellipse

Example 27 (IITianForum). Let $M$ and $N$ be the end points of that diameter of the circle $x^2+y^2-6x+6y-7=0$ which lies along the line $7x-y=24$. The tangents to an ellipse at the two end points of its major axis pass through $M$ and through $N$ respectively, and the line $MN$ itself touches the same ellipse. The major axis of the ellipse lies along the line $y=x$. Then

▸ SHOW SOLUTION

Step 1. M and N

The circle has centre $(3,-3)$ (on $7x-y=24$) and radius $5$. Unit vector along the line: $\frac{(1,7)}{5\sqrt2}$.

$$M,N=\left(3\pm\frac1{\sqrt2},\,-3\pm\frac7{\sqrt2}\right)$$

So $x+y=\pm4\sqrt2$ at $M$ and $N$.

Step 2. Centre and a

The vertex tangents are perpendicular to $y=x$, so they are $x+y=\pm4\sqrt2$. The centre lies midway on $y=x$: the origin. The distance between them is $8$, so $a=4$.

Step 3. Rotated coordinates

Take $X=\frac{x+y}{\sqrt2}$ (along the major axis), $Y=\frac{y-x}{\sqrt2}$. The ellipse is $\frac{X^2}{16}+\frac{Y^2}{b^2}=1$ and the line $7x-y=24$ becomes

$$3X-4Y=12\sqrt2$$

Step 4. Tangency

For $lX+mY=n$ to touch, $n^2=a^2l^2+b^2m^2$:

$$288=144+16b^2,\qquad b^2=9$$

Step 5. Check the options

$e^2=1-\frac9{16}$, so $e=\frac{\sqrt7}4$. Latus rectum $\frac{2\cdot9}4=\frac92$. The centre is $(0,0)$, so (C) fails. Directrix $X=\frac ae=\frac{16}{\sqrt7}$:

$$x+y=\frac{16\sqrt2}{\sqrt7}=\frac{16\sqrt{14}}7$$

Answer: (A), (B) and (D)

Example 28 (JEE Main 2026, 23 January Shift 2). If the points of intersection of the ellipses $x^2+2y^2-6x-12y+23=0$ and $4x^2+2y^2-20x-12y+35=0$ lie on a circle of radius $r$ and centre $(a,b)$, then the value of $ab+18r^2$ is

▸ SHOW SOLUTION

Step 1. Curve through the common points

$S_1+\lambda S_2=0$ passes through every common point. It is a circle when the $x^2$ and $y^2$ coefficients are equal:

$$1+4\lambda=2+2\lambda,\qquad\lambda=\frac12$$

Step 2. The circle

$$3x^2+3y^2-16x-18y+\frac{81}2=0$$$$x^2+y^2-\frac{16}3x-6y+\frac{27}2=0$$

Step 3. Centre and radius

Centre $\left(\frac83,3\right)$:

$$r^2=\frac{64}9+9-\frac{27}2=\frac{47}{18}$$$$ab+18r^2=8+47=55$$

Answer: option (4)

Position of a point and a line meeting the ellipse

Position of a point

For the ellipse $S\equiv\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$ write

$$S_1=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}-1$$

A line meeting the ellipse

Put $y=mx+c$ into the ellipse:

$$\left(b^2+a^2m^2\right)x^2+2a^2mcx$$ $$+a^2\left(c^2-b^2\right)=0$$

Its discriminant is $4a^2b^2\left(a^2m^2+b^2-c^2\right)$, so

Chord through a point in a given direction

The line through $P(x_1,y_1)$ at angle $\alpha$ is $x=x_1+r\cos\alpha$, $y=y_1+r\sin\alpha$. Putting it in the ellipse gives a quadratic in $r$ whose roots are $PA$ and $PB$ (with sign):

$$PA\cdot PB=\frac{|S_1|}{\dfrac{\cos^2\alpha}{a^2}+\dfrac{\sin^2\alpha}{b^2}}$$

Example 29 (IITianForum). For the ellipse $\frac{x^2}{16}+\frac{y^2}9=1$: (a) state whether $(2,2)$ and $(3,3)$ are inside or outside; (b) find the values of $c$ for which $y=x+c$ cuts the ellipse at two points, and the point of contact when it touches with $c>0$.

▸ SHOW SOLUTION

Step 1. Points

$$S_1(2,2)=\frac4{16}+\frac49-1=-\frac{11}{36}<0$$$$S_1(3,3)=\frac9{16}+1-1=\frac9{16}>0$$

So $(2,2)$ is inside and $(3,3)$ is outside.

Step 2. Line

With $m=1$: two points when $c^2<16+9$, that is $-5<c<5$.

Step 3. Contact for c = 5

The quadratic becomes $25x^2+160x+256=0$, that is $(5x+16)^2=0$:

$$x=-\frac{16}5,\qquad y=\frac95$$

Answer: $(2,2)$ inside, $(3,3)$ outside; $-5<c<5$; contact $\left(-\frac{16}5,\frac95\right)$

Example 30 (JEE Main 2026, 23 January Shift 1). Let the line $y-x=1$ intersect the ellipse $\frac{x^2}2+\frac{y^2}1=1$ at the points $A$ and $B$. Then the angle made by the line segment $AB$ at the centre of the ellipse is

▸ SHOW SOLUTION

Step 1. The points

$$x^2+2(x+1)^2=2\ \Rightarrow\ 3x^2+4x=0$$

$A(0,1)$ and $B\left(-\frac43,-\frac13\right)$.

Step 2. Angle at O

$\vec{OA}=(0,1)$ and $|OB|=\frac{\sqrt{17}}3$:

$$\cos\angle AOB=\frac{-\frac13}{\frac{\sqrt{17}}3}=-\frac1{\sqrt{17}}$$

Step 3. Write it with tan

$\cos\left(\frac\pi2+t\right)=-\sin t$ with $\sin t=\frac1{\sqrt{17}}$, $\tan t=\frac14$.

Answer: option (2)

Example 31 (JEE Main 2025, 3 April Shift 1). A line passing through the point $P\left(\sqrt5,\sqrt5\right)$ intersects the ellipse $\frac{x^2}{36}+\frac{y^2}{25}=1$ at $A$ and $B$ such that $(PA)\cdot(PB)$ is maximum. Then $5\left(PA^2+PB^2\right)$ is equal to

▸ SHOW SOLUTION

Step 1. P is inside

$$S_1=\frac5{36}+\frac5{25}-1=-\frac{595}{900}<0$$

Step 2. The product

$$PA\cdot PB=\frac{595}{25\cos^2\alpha+36\sin^2\alpha}$$

The denominator is least at $\alpha=0$, so the chord is horizontal: $y=\sqrt5$.

Step 3. The ends

$$\frac{x^2}{36}=1-\frac15,\qquad x=\pm\frac{12}{\sqrt5}$$

Step 4. Evaluate

$$PA^2+PB^2=2\left(5+\frac{144}5\right)=\frac{338}5$$

$5\left(PA^2+PB^2\right)=338$.

Answer: option (4)

Chords of the ellipse

Chord joining two points

The chord joining the points $\theta$ and $\phi$ is

$$\frac xa\cos\frac{\theta+\phi}2+\frac yb\sin\frac{\theta+\phi}2=\cos\frac{\theta-\phi}2$$

To see it, check that $(a\cos\theta,b\sin\theta)$ satisfies it: the left side is $\cos\theta\cos\frac{\theta+\phi}2+\sin\theta\sin\frac{\theta+\phi}2=\cos\frac{\theta-\phi}2$. The same holds for $\phi$.

Focal chords

The chord $\theta\phi$ passes through $S(ae,0)$ when $e\cos\frac{\theta+\phi}2=\cos\frac{\theta-\phi}2$, which simplifies to

$$\tan\frac\theta2\tan\frac\phi2=\frac{e-1}{e+1}$$

For the other focus $S'(-ae,0)$ the right side is $\frac{e+1}{e-1}$.

An ellipse with a focal chord PQ through the focus S, the eccentric angles of P and Q marked by their points on the auxiliary circle

If $PQ$ is a focal chord through $S$, the semi-latus rectum is the harmonic mean of $SP$ and $SQ$:

$$\frac1{SP}+\frac1{SQ}=\frac{2a}{b^2}$$

Chord with a given middle point

The chord of the ellipse whose middle point is $(x_1,y_1)$ is

$$T=S_1$$

where $T=\frac{xx_1}{a^2}+\frac{yy_1}{b^2}-1$. Its slope is $-\frac{b^2x_1}{a^2y_1}$.

An ellipse with a chord AB and its middle point M marked, the line CM joining the centre to M drawn dashed

Diameters and conjugate diameters

The middle points of all chords of slope $m$ lie on the line through the centre

$$y=-\frac{b^2}{a^2m}x$$

called a diameter. Two diameters $y=m_1x$ and $y=m_2x$ are conjugate (each bisects the chords parallel to the other) when

$$m_1m_2=-\frac{b^2}{a^2}$$

If $CP$ and $CD$ are conjugate semi-diameters with $P$ the point $\theta$, then $D$ is the point $\theta+\frac\pi2$, and

An ellipse with two conjugate semi-diameters CP and CD, and the parallelogram formed by the tangents at the four ends

Example 32 (IITianForum). For the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$, the focal chord through $B(0,4)$ and the focus $S(3,0)$ meets the ellipse again at $Q$. Find $Q$, check $\tan\frac\theta2\tan\frac\phi2=\frac{e-1}{e+1}$ and find the length of the chord.

▸ SHOW SOLUTION

Step 1. The chord

$e=\frac35$. Line $BS$: $y=4-\frac43x$. Put in the ellipse:

$$\frac{x^2}{25}+\left(1-\frac x3\right)^2=1\ \Rightarrow\ \frac{34x^2}{225}=\frac{2x}3$$

$x=\frac{75}{17}$, $y=-\frac{32}{17}$.

Step 2. Eccentric angles

$B$ is $\theta=\frac\pi2$, so $\tan\frac\theta2=1$. For $Q$: $\cos\phi=\frac{15}{17}$, $\sin\phi=-\frac8{17}$:

$$\tan\frac\phi2=\frac{\sin\phi}{1+\cos\phi}=-\frac14$$

And $\frac{e-1}{e+1}=\frac{-2/5}{8/5}=-\frac14$. The product agrees.

Step 3. Length

$SB=5$ and $SQ=5-\frac35\cdot\frac{75}{17}=\frac{40}{17}$:

$$SB+SQ=5+\frac{40}{17}=\frac{125}{17}$$

(Check: $\frac15+\frac{17}{40}=\frac58=\frac{2a}{b^2}$.)

Answer: $Q\left(\frac{75}{17},-\frac{32}{17}\right)$, length $\frac{125}{17}$

Example 33 (JEE Main 2026, 6 April Shift 2). Let $x=9$ be a directrix of an ellipse $E$, whose centre is at the origin and eccentricity is $\frac13$. Let $P(\alpha,0)$, $\alpha>0$, be a focus of $E$ and $AB$ be a chord passing through $P$. Then the locus of the mid-point of $AB$ is

▸ SHOW SOLUTION

Step 1. The ellipse

$\frac ae=9$, $e=\frac13$: $a=3$, $ae=1$, $b^2=8$. So $P=(1,0)$.

Step 2. Chord with middle point (h, k)

$$\frac{hx}9+\frac{ky}8=\frac{h^2}9+\frac{k^2}8$$

Step 3. Through (1, 0)

$$\frac h9=\frac{h^2}9+\frac{k^2}8\ \Rightarrow\ 8h=8h^2+9k^2$$

The locus is $9y^2=8x(1-x)$.

Answer: option (1)

Example 34 (JEE Main 2025, 23 January Shift 2). The length of the chord of the ellipse $\frac{x^2}4+\frac{y^2}2=1$, whose mid-point is $\left(1,\frac12\right)$, is

▸ SHOW SOLUTION

Step 1. The chord

$T=S_1$:

$$\frac x4+\frac y4=\frac14+\frac18,\qquad x+y=\frac32$$

Step 2. Meet the ellipse

$$x^2+2\left(\frac32-x\right)^2=4$$$$6x^2-12x+1=0$$$$|x_1-x_2|=\frac{\sqrt{144-24}}6=\frac{\sqrt{30}}3$$

Step 3. Length

The slope is $-1$, so the length is $\sqrt2\,|x_1-x_2|$:

$$\frac{\sqrt{60}}3=\frac23\sqrt{15}$$

Answer: option (1)

Tangents

The three forms of the tangent

The point form is $T=0$. It comes from the slope $-\frac{b^2x_1}{a^2y_1}$ found by differentiating $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. The slope form is the condition $c^2=a^2m^2+b^2$ found earlier; matching it with the point form gives the point of contact.

An ellipse with a tangent and the normal at a point P, the normal meeting the major axis at G and the minor axis at g, and the foot F of the perpendicular from the centre to the normal

Example 35 (IITianForum). For the ellipse $\frac{x^2}9+\frac{y^2}4=1$, find (a) the tangent at $\left(\frac95,\frac85\right)$; (b) the tangents parallel to $y=x$ and their points of contact.

▸ SHOW SOLUTION

Step 1. Point form

$\frac{81}{225}+\frac{64}{100}=\frac9{25}+\frac{16}{25}=1$, so the point is on the ellipse. Tangent:

$$\frac{9x}{45}+\frac{8y}{20}=1,\qquad x+2y=5$$

Step 2. Slope form

$m=1$: $c=\pm\sqrt{9+4}=\pm\sqrt{13}$.

Step 3. Points of contact

For $y=x+\sqrt{13}$: $\left(-\frac9{\sqrt{13}},\frac4{\sqrt{13}}\right)$. For $y=x-\sqrt{13}$: $\left(\frac9{\sqrt{13}},-\frac4{\sqrt{13}}\right)$.

Answer: $x+2y=5$; $y=x\pm\sqrt{13}$ at $\left(\mp\frac9{\sqrt{13}},\pm\frac4{\sqrt{13}}\right)$

Example 36 (JEE Main 2026, 21 January Shift 2). The line $\alpha x+4y=\sqrt7$, $\alpha\in\mathbb{R}$, touches the ellipse $3x^2+4y^2=1$ at the point $P$ in the first quadrant. Then one of the focal distances of $P$ is

▸ SHOW SOLUTION

Step 1. Compare with the point form

The tangent at $(x_1,y_1)$ is $3x_1x+4y_1y=1$. Matching with $\alpha x+4y=\sqrt7$:

$$y_1=\frac1{\sqrt7},\qquad3x_1=\frac\alpha{\sqrt7}$$

Step 2. Find P

$3x_1^2+\frac47=1$ gives $x_1^2=\frac17$; first quadrant, so $x_1=\frac1{\sqrt7}$.

Step 3. Focal distances

$a^2=\frac13$, $b^2=\frac14$, $e^2=1-\frac34=\frac14$:

$$a\pm ex_1=\frac1{\sqrt3}\pm\frac1{2\sqrt7}$$

Answer: option (4)

Example 37 (JEE Main 2025, 4 April Shift 2). Let for two distinct values of $p$ the lines $y=x+p$ touch the ellipse $E:\frac{x^2}{4^2}+\frac{y^2}{3^2}=1$ at the points $A$ and $B$. Let the line $y=x$ intersect $E$ at the points $C$ and $D$. Then the area of the quadrilateral $ABCD$ is equal to

▸ SHOW SOLUTION

Step 1. A and B

$p^2=16+9$, $p=\pm5$. Contacts $\left(\mp\frac{16}5,\pm\frac95\right)$.

Step 2. C and D

$x^2\left(\frac1{16}+\frac19\right)=1$ gives $x=\pm\frac{12}5$: $\pm\left(\frac{12}5,\frac{12}5\right)$.

Step 3. Area

Both pairs are symmetric about the origin, so $ACBD$ is a parallelogram with diagonals $\vec{AB}=\left(\frac{32}5,-\frac{18}5\right)$ and $\vec{CD}=\left(\frac{24}5,\frac{24}5\right)$:

$$\frac12\left|\frac{32\cdot24+18\cdot24}{25}\right|=24$$

Answer: option (2)

Where two tangents meet

The tangents at the points $\theta$ and $\phi$ meet at

$$\left(\frac{a\cos\frac{\theta+\phi}2}{\cos\frac{\theta-\phi}2},\ \frac{b\sin\frac{\theta+\phi}2}{\cos\frac{\theta-\phi}2}\right)$$

Pair of tangents and chord of contact

From $P(x_1,y_1)$ outside the ellipse two tangents can be drawn.

An ellipse and an outside point P(3, 4) with the two tangents from P touching at A(3, 0) and B, and the chord of contact AB

Director circle

The locus of the point from which the two tangents are perpendicular is the director circle

$$x^2+y^2=a^2+b^2$$

Reason: $y=mx+c$ through $(h,k)$ with $c^2=a^2m^2+b^2$ gives $\left(h^2-a^2\right)m^2-2hkm+k^2-b^2=0$. Perpendicular tangents need the product of the roots $-1$:

$$k^2-b^2=-\left(h^2-a^2\right)$$ An ellipse with its larger concentric director circle, a point T on the circle and the two perpendicular tangents from T to the ellipse

Pole and polar

The polar of $P(x_1,y_1)$ with respect to the ellipse is $T=0$ (for $P$ outside, it is the chord of contact). The pole of the line $lx+my+n=0$ is

$$\left(-\frac{a^2l}n,-\frac{b^2m}n\right)$$

If the polar of $P$ passes through $Q$, then the polar of $Q$ passes through $P$ (conjugate points). The polar of a focus is the corresponding directrix, so tangents at the ends of a focal chord meet on the directrix.

Standard properties of a tangent

The first line comes from $\frac{|c|}{\sqrt{1+m^2}}=\sqrt{\frac{a^2m^2+b^2}{1+m^2}}$, a weighted mean of $a^2$ and $b^2$.

An ellipse with a tangent, the perpendiculars from both foci to the tangent, and their feet lying on the auxiliary circle

Example 38 (IIT JEE 2010, Paper 2). Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}9+\frac{y^2}4=1$ touching the ellipse at points $A$ and $B$. The coordinates of $A$ and $B$ are

▸ SHOW SOLUTION

Step 1. Chord of contact

$$\frac{3x}9+\frac{4y}4=1,\qquad x=3-3y$$

Step 2. Meet the ellipse

$$(1-y)^2+\frac{y^2}4=1\ \Rightarrow\ \frac54y^2-2y=0$$

$y=0$ or $y=\frac85$.

Step 3. The points

$(3,0)$ and $\left(3-\frac{24}5,\frac85\right)=\left(-\frac95,\frac85\right)$.

Answer: option (D)

Example 39 (IIT JEE 2010, Paper 2). For $P$, $A$, $B$ of Example 38, the orthocentre of the triangle $PAB$ is

▸ SHOW SOLUTION

Step 1. One altitude

$PA$ is the vertical line $x=3$, so the altitude from $B$ is horizontal: $y=\frac85$.

Step 2. Second altitude

Slope of $PB=\frac{4-\frac85}{3+\frac95}=\frac12$. The altitude from $A(3,0)$ has slope $-2$:

$$y=-2x+6$$

Step 3. Meet

$y=\frac85$ gives $x=\frac{11}5$.

Answer: option (C)

Example 40 (JEE Main 2023, 1 February Shift 2). The line $x=8$ is the directrix of the ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with the corresponding focus $(2,0)$. If the tangent to $E$ at the point $P$ in the first quadrant passes through the point $\left(0,4\sqrt3\right)$ and intersects the $x$-axis at $Q$, then $(3PQ)^2$ is equal to ______.

▸ SHOW SOLUTION

Step 1. The ellipse

$\frac ae=8$, $ae=2$: $a^2=16$, $e=\frac12$, $b^2=12$.

Step 2. The point P

The tangent $\frac{xx_1}{16}+\frac{yy_1}{12}=1$ passes through $\left(0,4\sqrt3\right)$: $y_1=\sqrt3$. Then $x_1=2\sqrt3$.

Step 3. Q and PQ

$y=0$ gives $x=\frac{16}{x_1}=\frac8{\sqrt3}$.

$$PQ^2=\left(\frac8{\sqrt3}-2\sqrt3\right)^2+3$$$$=\frac43+3=\frac{13}3$$

$(3PQ)^2=9\cdot\frac{13}3=39$.

Answer: $39$

An ellipse tilted in the first quadrant touching both coordinate axes, its centre on an arc of the circle of radius 2 about the origin

Example 41 (IITianForum). An ellipse whose major axis has length $2\sqrt3$ and whose minor axis has length $2$ slides along the coordinate axes, touching both of them, and stays inside the first quadrant throughout. The locus of its centre is an arc of a circle. The length of this arc is

▸ SHOW SOLUTION

Step 1. The circle

The two axes are perpendicular tangents, so the origin lies on the director circle. Hence the centre is at distance $\sqrt{a^2+b^2}=\sqrt{3+1}=2$ from $O$.

Step 2. Range of the centre

The ordinate $k$ of the centre is its distance from the tangent $y=0$, which lies between $b=1$ and $a=\sqrt3$. With the centre at angle $\psi$ on the circle, $k=2\sin\psi$:

$$\frac12\le\sin\psi\le\frac{\sqrt3}2,\qquad\frac\pi6\le\psi\le\frac\pi3$$

Step 3. Arc length

$$2\left(\frac\pi3-\frac\pi6\right)=\frac\pi3$$

Answer: option (B)

Example 42 (IITianForum). From a point $T$ on the line $12x+5y=169$, two mutually perpendicular tangents are drawn to the ellipse $\frac{x^2}{144}+\frac{y^2}{25}=1$, touching it at $A$ and $B$. The normals at $A$ and $B$ meet at $P$. The area of the quadrilateral $TAPB$ is ______.

▸ SHOW SOLUTION

Step 1. T lies on the director circle

Perpendicular tangents meet on $x^2+y^2=169$. The distance of the line from $O$ is $\frac{169}{13}=13$, so the line touches this circle, at $T(12,5)$ only.

Step 2. The tangents

From $(12,5)$: $x=12$ touches at $A(12,0)$ and $y=5$ touches at $B(0,5)$. They are perpendicular.

Step 3. The normals

The normals at the vertices $A$ and $B$ are the axes $y=0$ and $x=0$, so $P=(0,0)$.

Step 4. Area

$TAPB$ is a rectangle with sides $12$ and $5$: area $60$.

Answer: $60$

Common tangents

A common tangent $y=mx+c$ to two curves satisfies both tangency conditions. For two ellipses this gives two equations in $m^2$ and $c^2$.

Example 43 (JEE Advanced 2015, Paper 2). Let $E_1$ and $E_2$ be two ellipses whose centres are at the origin. The major axes of $E_1$ and $E_2$ lie along the $x$-axis and the $y$-axis, respectively. Let $S$ be the circle $x^2+(y-1)^2=2$. The straight line $x+y=3$ touches the curves $S$, $E_1$ and $E_2$ at $P$, $Q$ and $R$, respectively. Suppose that $PQ=PR=\frac{2\sqrt2}3$. If $e_1$ and $e_2$ are the eccentricities of $E_1$ and $E_2$, respectively, then the correct expression(s) is (are)

▸ SHOW SOLUTION

Step 1. P

The foot of the perpendicular from $(0,1)$ to $x+y=3$ is $P(1,2)$.

Step 2. Q and R

Moving $\frac{2\sqrt2}3$ along the line (direction $\frac{(1,-1)}{\sqrt2}$) from $P$: $\left(\frac53,\frac43\right)$ and $\left(\frac13,\frac83\right)$.

Step 3. Match with the point form

The tangent at $(x_1,y_1)$ is $\frac{xx_1}{p^2}+\frac{yy_1}{q^2}=1$; it is $x+y=3$ when $p^2=3x_1$, $q^2=3y_1$. $E_1$ needs $p>q$, so $Q=\left(\frac53,\frac43\right)$:

$$E_1:\ p^2=5,\ q^2=4,\qquad e_1^2=\frac15$$$$E_2:\ p^2=1,\ q^2=8,\qquad e_2^2=\frac78$$

Step 4. Check

$e_1^2+e_2^2=\frac{43}{40}$, $e_1e_2=\sqrt{\frac7{40}}=\frac{\sqrt7}{2\sqrt{10}}$, $\left|e_1^2-e_2^2\right|=\frac{27}{40}$.

Answer: (A) and (B)

Example 44 (IITianForum). The ellipses $\frac{x^2}9+y^2=1$, $\frac{x^2}4+\frac{y^2}k=1$ and $x^2+\frac{y^2}4=1$ have a common tangent. The value of $8k$ is ______.

▸ SHOW SOLUTION

Step 1. Three conditions

For $y=mx+c$:

$$c^2=9m^2+1=4m^2+k=m^2+4$$

Step 2. First and third

$9m^2+1=m^2+4$: $m^2=\frac38$, $c^2=\frac{35}8$.

Step 3. Middle one

$$k=c^2-4m^2=\frac{35}8-\frac{12}8=\frac{23}8$$

(In general, $c^2=a_i^2m^2+b_i^2$ for three ellipses forces the determinant with rows $\left(a_i^2,b_i^2,1\right)$ to be $0$.)

Answer: $23$

Example 45 (IITianForum). A tangent of slope $\frac12$ of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ $(a>b)$ is normal to the circle $x^2+y^2-2x+2y+1=0$. Then

▸ SHOW SOLUTION

Step 1. The tangent

A normal to the circle passes through its centre $(1,-1)$. So $y=\frac x2+c$ with $c=-\frac32$.

Step 2. Tangency

$$\frac94=\frac{a^2}4+b^2,\qquad a^2+4b^2=9$$

Step 3. Range of a

$b^2>0$ gives $a<3$; $b<a$ gives $9-a^2<4a^2$, $a>\frac3{\sqrt5}$. So (B) holds; (C) fails ($a=3$ gives $b=0$).

Step 4. Largest ab

$$a^2b^2=\frac{a^2\left(9-a^2\right)}4\le\frac14\cdot\frac{81}4$$

by AM-GM, with equality at $a^2=\frac92$, inside the range. So the largest $ab$ is $\frac94$.

Answer: (A) and (B)

Example 46 (JEE Main 2020, 6 September Shift 1). Which of the following points lies on the locus of the foot of perpendicular drawn upon any tangent to the ellipse $\frac{x^2}4+\frac{y^2}2=1$ from any of its foci?

▸ SHOW SOLUTION

Step 1. The locus

The feet lie on the auxiliary circle $x^2+y^2=4$.

Step 2. Test the points

$1+4=5$, $4+3=7$, $1+3=4$, $1+2=3$. Only $\left(-1,\sqrt3\right)$ lies on it.

Answer: option (3)

Example 47 (JEE Main 2021, 25 July Shift 2). If a tangent to the ellipse $x^2+4y^2=4$ meets the tangents at the extremities of its major axis at $B$ and $C$, then the circle with $BC$ as diameter passes through the point

▸ SHOW SOLUTION

Step 1. B and C

Tangent at $\theta$: $\frac x2\cos\theta+y\sin\theta=1$. At $x=\pm2$:

$$y_B=\frac{1-\cos\theta}{\sin\theta},\qquad y_C=\frac{1+\cos\theta}{\sin\theta}$$

Step 2. Circle on BC

$$(x-2)(x+2)+(y-y_B)(y-y_C)=0$$

Step 3. Put y = 0

$y_By_C=1$, so $x^2-4+1=0$, $x=\pm\sqrt3$. These are the foci, for every $\theta$.

Answer: option (1)

Example 48 (JEE Main 2021, 26 August Shift 1). On the ellipse $\frac{x^2}8+\frac{y^2}4=1$ let $P$ be a point in the second quadrant such that the tangent at $P$ to the ellipse is perpendicular to the line $x+2y=0$. Let $S$ and $S'$ be the foci of the ellipse and $e$ be its eccentricity. If $A$ is the area of the triangle $SPS'$, then the value of $\left(5-e^2\right)\cdot A$ is

▸ SHOW SOLUTION

Step 1. Slope form

Slope $2$: $c=\pm\sqrt{8\cdot4+4}=\pm6$. Contact for $c=6$:

$$\left(-\frac{8\cdot2}6,\frac46\right)=\left(-\frac83,\frac23\right)$$

This is in the second quadrant.

Step 2. Area

$e^2=\frac12$, foci $(\pm2,0)$, $SS'=4$:

$$A=\frac12\cdot4\cdot\frac23=\frac43$$

Step 3. Evaluate

$$\left(5-\frac12\right)\cdot\frac43=6$$

Answer: option (1)

Normals

The three forms of the normal

In the slope form the foot is $\left(\pm\frac{a^2}{\sqrt{a^2+b^2m^2}},\pm\frac{b^2m}{\sqrt{a^2+b^2m^2}}\right)$ (same order of signs). So $y=mx+c$ is a normal when

$$c^2=\frac{m^2\left(a^2-b^2\right)^2}{a^2+b^2m^2}$$

The normal and the axes

The normal at $P(x_1,y_1)$ meets the major axis at $G$ and the minor axis at $g$:

$$G\left(e^2x_1,0\right),\qquad g\left(0,-\frac{a^2e^2}{b^2}y_1\right)$$

If $F$ is the foot of the perpendicular from the centre to the normal at $P$, then

$$PF\cdot PG=b^2,\qquad PF\cdot Pg=a^2$$

So $PG:Pg=b^2:a^2$, and $CG=e^2x_1$ is always less than $ae$: the normal meets the major axis between the foci.

Co-normal points

The normals at $(a\cos\theta,b\sin\theta)$ through $(h,k)$ satisfy

$$\frac{ah}{\cos\theta}-\frac{bk}{\sin\theta}=a^2-b^2$$

which is a fourth degree equation in $\tan\frac\theta2$. So at most four normals pass through a point; their feet are co-normal points.

The last condition comes from the determinant with rows $\left(\sec\alpha,\text{cosec}\,\alpha,1\right)$ being $0$.

An ellipse with a point inside it and the four normals from that point meeting the ellipse at four co-normal points

Example 49 (IITianForum). For the ellipse $\frac{x^2}{25}+\frac{y^2}9=1$, find the normal at the point $\theta=\frac\pi4$, the points $G$ and $g$ where it meets the axes, and check that $PG:Pg=9:25$.

▸ SHOW SOLUTION

Step 1. The normal

$P\left(\frac5{\sqrt2},\frac3{\sqrt2}\right)$:

$$5\sqrt2\,x-3\sqrt2\,y=16$$

Step 2. G and g

$$G\left(\frac{8\sqrt2}5,0\right),\qquad g\left(0,-\frac{8\sqrt2}3\right)$$

($CG=e^2x_1=\frac{16}{25}\cdot\frac5{\sqrt2}$, as expected.)

Step 3. The ratio

$$PG^2=\frac{81}{50}+\frac92=\frac{306}{50}$$$$Pg^2=\frac{25}2+\frac{625}{18}=\frac{850}{18}$$

$\frac{PG^2}{Pg^2}=\frac{81}{625}$, so $PG:Pg=9:25$.

Answer: $5\sqrt2x-3\sqrt2y=16$, $G\left(\frac{8\sqrt2}5,0\right)$, $g\left(0,-\frac{8\sqrt2}3\right)$

Example 50 (JEE Main 2020, 6 September Shift 2). If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity $e$ of the ellipse satisfies

▸ SHOW SOLUTION

Step 1. The normal at L

$L\left(ae,\frac{b^2}a\right)$:

$$\frac{a^2x}{ae}-\frac{b^2y}{b^2/a}=a^2e^2$$$$\frac xe-y=ae^2$$

Step 2. Through (0, -b)

$b=ae^2$, so $b^2=a^2e^4$.

Step 3. Use b squared

$$a^2\left(1-e^2\right)=a^2e^4,\qquad e^4+e^2-1=0$$

Answer: option (4)

Example 51 (IIT JEE 2009, Paper 2). The normal at a point $P$ on the ellipse $x^2+4y^2=16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $PQ$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points

▸ SHOW SOLUTION

Step 1. P and Q

$\frac{x^2}{16}+\frac{y^2}4=1$, $e^2=\frac34$. $P(4\cos\theta,2\sin\theta)$ and $Q\left(e^2\cdot4\cos\theta,0\right)=(3\cos\theta,0)$.

Step 2. M

$$M=\left(\frac72\cos\theta,\ \sin\theta\right)$$

Locus: $\frac{4x^2}{49}+y^2=1$.

Step 3. Meet the latus rectum

$x=\pm ae=\pm2\sqrt3$:

$$y^2=1-\frac{48}{49}=\frac1{49}$$

Answer: option (C)

Example 52 (JEE Main 2023, 31 January Shift 1). If the maximum distance of normal to the ellipse $\frac{x^2}4+\frac{y^2}{b^2}=1$, $b<2$, from the origin is $1$, then the eccentricity of the ellipse is

▸ SHOW SOLUTION

Step 1. Distance of the normal at θ

$$d=\frac{a^2-b^2}{\sqrt{a^2\sec^2\theta+b^2\text{cosec}^2\theta}}$$

Step 2. Least denominator

$$a^2\sec^2\theta+b^2\text{cosec}^2\theta$$$$=a^2+b^2+a^2\tan^2\theta+b^2\cot^2\theta$$

By AM-GM, $a^2\tan^2\theta+b^2\cot^2\theta\ge2ab$, so the sum is at least $(a+b)^2$.

So the largest $d$ is $\frac{a^2-b^2}{a+b}=a-b$.

Step 3. Use d = 1

$2-b=1$, $b=1$:

$$e^2=1-\frac14,\qquad e=\frac{\sqrt3}2$$

Answer: option (2)

Example 53 (IIT JEE 2009, Paper 2). An ellipse intersects the hyperbola $2x^2-2y^2=1$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

▸ SHOW SOLUTION

Step 1. Eccentricities

The hyperbola $x^2-y^2=\frac12$ has $e=\sqrt2$, so the ellipse has $e^2=\frac12$: $b^2=\frac{a^2}2$, that is $x^2+2y^2=a^2$.

Step 2. Orthogonal at a common point

Slopes: ellipse $-\frac x{2y}$, hyperbola $\frac xy$. Product $-1$:

$$x^2=2y^2$$

Step 3. The point and a

On the hyperbola: $2y^2-y^2=\frac12$, so $y^2=\frac12$, $x^2=1$. Then $a^2=1+1=2$.

Step 4. Foci

$x^2+2y^2=2$: $a^2=2$, $b^2=1$, $ae=1$, foci $(\pm1,0)$.

Answer: (A) and (B)

Reflection property, area and standard results

Reflection property

The tangent at $P$ makes equal angles with $SP$ and $S'P$; the normal at $P$ bisects the angle $SPS'$. So a ray from one focus, reflected at the ellipse, passes through the other focus.

An ellipse with a ray from the focus S' striking the curve at P and reflecting through the other focus S, the normal at P bisecting the angle SPS'

Area of an ellipse

The ellipse is the auxiliary circle compressed in the ratio $b:a$, so every area shrinks in that ratio:

$$\text{Area}=\pi a^2\cdot\frac ba=\pi ab$$

More standard results

Example 54 (IITianForum). A circle of radius $r$ passes through both foci of the ellipse $x^2+9y^2=9$ and through exactly four points of the ellipse. The set of all possible values of $r$ is an interval $[a,b)$. Then $a+b$ is equal to

▸ SHOW SOLUTION

Step 1. The circles

Foci $\left(\pm2\sqrt2,0\right)$. The centre lies on the $y$-axis, $(0,k)$:

$$x^2+y^2-2ky-8=0,\qquad r^2=8+k^2$$

Step 2. Meet the ellipse

Put $x^2=9-9y^2$:

$$8y^2+2ky-1=0$$

Each root with $|y|<1$ gives two points.

Step 3. Both roots in (-1, 1)

The product of roots is $-\frac18<0$, so they have opposite signs. Need $f(1)=7+2k>0$ and $f(-1)=7-2k>0$: $|k|<\frac72$.

Step 4. Range of r

$$r\in\left[2\sqrt2,\ \sqrt{8+\tfrac{49}4}\right)=\left[2\sqrt2,\frac92\right)$$

Answer: option (A)

Example 55 (JEE Main 2025, 4 April Shift 2). The centre of a circle $C$ is at the centre of the ellipse $E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, $a>b$. Let $C$ pass through the foci $F_1$ and $F_2$ of $E$ such that the circle $C$ and the ellipse $E$ intersect at four points. Let $P$ be one of these four points. If the area of the triangle $PF_1F_2$ is $30$ and the length of the major axis of $E$ is $17$, then the distance between the foci of $E$ is

▸ SHOW SOLUTION

Step 1. Right angle at P

$F_1F_2$ is a diameter of $C$, so $\angle F_1PF_2=90^\circ$. Let $PF_1=p$, $PF_2=q$.

Step 2. Two equations

$$p+q=17,\qquad\frac12pq=30$$

Step 3. Pythagoras

$$F_1F_2^2=p^2+q^2=289-120=169$$

Answer: option (2)

Example 56 (IITianForum). A ray of light from $(-4,0)$ strikes the mirror $\frac{x^2}{25}+\frac{y^2}9=1$ at $P(0,3)$. Find the equation of the reflected ray.

▸ SHOW SOLUTION

Step 1. The foci

$a=5$, $b=3$, $ae=4$: $(-4,0)$ is the focus $S'$.

Step 2. Reflection property

The reflected ray passes through the other focus $S(4,0)$.

Step 3. The line PS

$$\frac x4+\frac y3=1,\qquad3x+4y=12$$

Answer: $3x+4y=12$

Example 57 (JEE Main 2026, 21 January Shift 1). The area of the region inside the ellipse $x^2+4y^2=4$ and outside the region bounded by the curves $y=|x|-1$ and $y=1-|x|$ is

▸ SHOW SOLUTION

Step 1. The ellipse

$\frac{x^2}4+y^2=1$: area $\pi\cdot2\cdot1=2\pi$.

Step 2. The square

The two curves bound the square $|x|+|y|\le1$, with diagonals $2$ and $2$, area $2$. Its vertices $(\pm1,0)$, $(0,\pm1)$ are inside or on the ellipse.

Step 3. Subtract

$$2\pi-2=2(\pi-1)$$

Answer: option (1)

Example 58 (JEE Main 2025, 28 January Shift 1). Let $E_1:\frac{x^2}9+\frac{y^2}4=1$ be an ellipse. Ellipses $E_i$'s are constructed such that their centres and eccentricities are same as that of $E_1$, and the length of minor axis of $E_i$ is the length of major axis of $E_{i+1}$ $(i\ge1)$. If $A_i$ is the area of the ellipse $E_i$, then $\frac5\pi\left(\sum_{i=1}^{\infty}A_i\right)$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Ratio of sizes

Same eccentricity means the same ratio $\frac{\text{minor}}{\text{major}}=\frac23$. Each ellipse is the previous one scaled by $\frac23$.

Step 2. Areas

$A_1=6\pi$ and $A_{i+1}=\frac49A_i$:

$$\sum A_i=\frac{6\pi}{1-\frac49}=\frac{54\pi}5$$

Step 3. Evaluate

$$\frac5\pi\cdot\frac{54\pi}5=54$$

Answer: $54$

Loci connected with the ellipse

Method

Take the moving point as $(h,k)$, write the given condition using the eccentric angle (or $T=S_1$, or the chord of contact), eliminate the parameter, and replace $(h,k)$ by $(x,y)$.

Standard loci

Example 59 (IITianForum). A pair of variable straight lines $x^2+\alpha xy+9y^2=0$, where $\alpha$ is a real parameter, cuts the ellipse $x^2+9y^2=9$; let $A$ and $B$ be points of intersection, one on each line. The locus of the point of intersection of the tangents to the ellipse at $A$ and at $B$ is

▸ SHOW SOLUTION

Step 1. Chord of contact

If the tangents at $A$, $B$ meet at $(h,k)$, then $AB$ is $\frac{hx}9+ky=1$.

Step 2. Lines OA and OB

Make the ellipse homogeneous with this chord:

$$\frac{x^2}9+y^2=\left(\frac{hx}9+ky\right)^2$$$$\left(9-h^2\right)x^2-18hkxy$$$$+81\left(1-k^2\right)y^2=0$$

Step 3. Compare with the given pair

The ratio of the $y^2$ and $x^2$ coefficients must be $9$:

$$81-81k^2=81-9h^2,\qquad h^2=9k^2$$

Answer: option (A)

Example 60 (JEE Main 2021, 31 August Shift 2). The locus of mid-points of the line segments joining $(-3,-5)$ and the points on the ellipse $\frac{x^2}4+\frac{y^2}9=1$ is

▸ SHOW SOLUTION

Step 1. Middle point

$$h=\frac{2\cos\theta-3}2,\qquad k=\frac{3\sin\theta-5}2$$

Step 2. Eliminate

$\cos\theta=\frac{2h+3}2$, $\sin\theta=\frac{2k+5}3$:

$$\frac{(2h+3)^2}4+\frac{(2k+5)^2}9=1$$

Step 3. Expand

Multiply by $36$:

$$36h^2+16k^2+108h+80k+145=0$$

Answer: option (3)

Which method to use

A decision chart: from the question asked, arrows lead to situations paired with methods: sum of distances, foci and 2a; general form, complete the square; point on the curve, eccentric angle; focal chord, tan half angles; mid-point given, T equals S1; tangent, c squared equals a squared m squared plus b squared; perpendicular tangents, director circle; normal, the normal forms; locus, parametrise and eliminate

What JEE asks

Across the JEE Main papers from 2014 to 2026 and the JEE Advanced and IIT JEE papers, the chapter appears in these forms.

1. Parts of the ellipse: eccentricity, latus rectum, foci from given data, often with $a<b$ or a shifted centre (Examples 7 to 18 and 24 to 26). This is the most common form in JEE Main.
2. Focal distances and the sum definition, including complex number sets (Examples 2, 3, 10, 11, 13 and 55).
3. Eccentric angle and the auxiliary circle (Examples 19 to 23).
4. Chords: mid-point chords, focal chords, chords through a point (Examples 30 to 34).
5. Tangents: point and slope forms, chord of contact, director circle and common tangents (Examples 35 to 48). This is the common form in JEE Advanced.
6. Normals (Examples 49 to 53) and loci (Examples 59 and 60).

JEE Advanced 2025 used the auxiliary circle to read eccentric angles (Example 20). Earlier JEE Advanced and IIT JEE papers asked for the auxiliary circle and a chord (Example 19), the sum definition inside a triangle (Example 4), the rectangle round an ellipse (Example 18), tangents from a point (Examples 38 and 39), a common tangent of a circle and two ellipses (Example 43), the normal and a locus of mid-points (Example 51) and orthogonal curves (Example 53).

Quick recall

Practise these methods on real exam questions in the doMath Ellipse chapter. The parabola, the conic with $e=1$, is developed in the Parabola note, circles through the meeting points of two curves in the Family of Circles note, and tangents to a general curve in the Tangents, Normals, Rolle and LMVT note.