Hyperbola for JEE: Notes, Asymptotes, Tangents, Normals, xy = c² and PYQs

A hyperbola is the path of a comet that passes the sun once and never returns, the curve of constant time difference used in radio navigation, and the graph of $y=\frac kx$. In coordinate geometry it is the open conic with two branches: two foci, two directrices, eccentricity greater than $1$, and two asymptotes that the branches approach but never meet. Almost every formula of the ellipse carries over with $b^2$ replaced by $-b^2$, so the real work in this chapter is knowing exactly where a sign changes, and handling the two new objects: the asymptotes and the rectangular hyperbola $xy=c^2$.

This note covers the two definitions of the hyperbola, the standard hyperbola and its parts, focal distances, the conjugate hyperbola, the auxiliary circle and the point $(a\sec\theta,b\tan\theta)$, shifted hyperbolas and the general second degree equation, the rectangular hyperbola in both forms, the position of a point, a line meeting the hyperbola, 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, chords and focal chords, chords with a given middle point, diameters, asymptotes and their standard results, the curve $xy=c^2$, the reflection property 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 hyperbola

Let $S$ be a fixed point and $\ell$ a fixed line not through $S$. The locus of a point $P$ for which

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

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

So a hyperbola is the locus of a point whose distance from a fixed point is a fixed multiple $e>1$ of its distance from a fixed line:

$$PS=e\cdot PM,\qquad e>1$$ A hyperbola with focus S inside the right branch, the directrix as a dashed vertical line between the centre and the branch, a point P on the branch joined to S and to the foot M of its perpendicular on the directrix

The second definition: difference of focal distances

A hyperbola has two foci $S$ and $S'$. For every point $P$ on it,

$$\lvert PS-PS'\rvert=2a$$

where $2a$ is the length of the transverse axis. On the branch nearer $S$, $PS'-PS=2a$; on the other branch, $PS-PS'=2a$.

A hyperbola with both foci S and S' on the transverse axis and a point P on the right branch joined to both foci

The reason for the condition $k<SS'$: in triangle $PSS'$, the difference of two sides is less than the third side.

Example 1 (IITianForum). Find the locus of a point $P$ for which $\lvert PS-PS'\rvert=6$, where $S(5,0)$ and $S'(-5,0)$.

▸ SHOW SOLUTION

Step 1. Write the condition

For $P(x,y)$:

$$\sqrt{(x-5)^2+y^2}-\sqrt{(x+5)^2+y^2}=\pm6$$

Step 2. Square once

Write $R=\sqrt{(x+5)^2+y^2}$:

$$\sqrt{(x-5)^2+y^2}=R\pm6$$$$(x-5)^2+y^2=R^2\pm12R+36$$$$(x-5)^2=(x+5)^2+36\pm12R$$$$-20x-36=\pm12R$$$$5x+9=\mp3R$$

Step 3. Square again

$$25x^2+90x+81=9R^2$$$$=9x^2+90x+225+9y^2$$$$16x^2-9y^2=144$$

Step 4. Check with the formula

$2a=6$, $SS'=2ae=10$, so $a=3$, $ae=5$ and $b^2=25-9=16$.

Answer: $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$

Example 2 (IITianForum). Find the equation of the hyperbola with focus $(1,1)$, directrix $x+y=0$ and eccentricity $\sqrt2$. Show that its asymptotes are perpendicular.

▸ SHOW SOLUTION

Step 1. Focus directrix condition

$$PS^2=e^2\cdot PM^2$$$$(x-1)^2+(y-1)^2=2\cdot\frac{(x+y)^2}{2}$$

Step 2. Simplify

$$x^2+y^2-2x-2y+2=x^2+2xy+y^2$$$$xy+x+y=1$$$$(x+1)(y+1)=2$$

Step 3. Read the curve

With $X=x+1$, $Y=y+1$ it is $XY=2$, a hyperbola of the form $xy=c^2$ with centre $(-1,-1)$. Its asymptotes are $x=-1$ and $y=-1$, perpendicular. This matches $e=\sqrt2$, the eccentricity of every rectangular hyperbola.

Answer: $(x+1)(y+1)=2$, a rectangular hyperbola with centre $(-1,-1)$

Example 3 (JEE Main 2025, 22 January Shift 1). Let the foci of a hyperbola be $(1,14)$ and $(1,-12)$. If it passes through the point $(1,6)$, then the length of its latus rectum is

▸ SHOW SOLUTION

Step 1. Use the difference definition

$P(1,6)$ lies on the line of the foci:

$$PS=14-6=8$$$$PS'=6+12=18$$$$2a=\lvert18-8\rvert=10,\qquad a=5$$

Step 2. Find b

$SS'=26=2ae$, so $ae=13$:

$$b^2=a^2e^2-a^2=169-25=144$$

Step 3. Latus rectum

$$\frac{2b^2}{a}=\frac{288}{5}$$

Answer: option (3)

Recognising a hyperbola

For the general second degree equation

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

let

$$\Delta=abc+2fgh-af^2-bg^2-ch^2$$

With no $xy$ term: a hyperbola has $x^2$ and $y^2$ with opposite signs.

Example 4 (JEE Main 2019, 10 January Shift 2). Let

$$S=\left\{(x,y)\in\mathbb R^2:\frac{y^2}{1+r}-\frac{x^2}{1-r}=1\right\}$$

where $r\ne\pm1$. Then $S$ represents

▸ SHOW SOLUTION

Step 1. Case 0 < r < 1

Both denominators are positive, so $S$ is a hyperbola with transverse axis along the $y$-axis:

$$e^2=1+\frac{1-r}{1+r}=\frac{2}{1+r}$$

$e=\sqrt{\frac{2}{1+r}}$, which matches neither (1) nor (3).

Step 2. Case r > 1

$1-r<0$, so

$$\frac{y^2}{r+1}+\frac{x^2}{r-1}=1$$

an ellipse with $r+1>r-1$, major axis along the $y$-axis:

$$e^2=1-\frac{r-1}{r+1}=\frac{2}{r+1}$$

Answer: option (2)

The standard hyperbola

The hyperbola with centre at the origin and foci on the $x$-axis is

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\qquad b^2=a^2\left(e^2-1\right)$$

Here $a$ and $b$ are just positive numbers: $a>b$, $a=b$ and $a<b$ are all allowed. Which axis is transverse is decided by the sign, not by the size of the denominators.

Parts of the hyperbola

The standard hyperbola with centre C, vertices A and A', foci S and S', both directrices dashed, the latus rectum LL' through S, the ends B and B' of the conjugate axis, and the two asymptotes as faint dashed lines

The points $B$, $B'$ do not lie on the curve; the conjugate axis only fixes $b$.

Ellipse and hyperbola side by side

Example 5 (JEE Main 2026, 21 January Shift 1). Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$. If the eccentricity of the hyperbola is $5$, then the length of its latus rectum is

▸ SHOW SOLUTION

Step 1. Foci of the ellipse

$$c^2=36-16=20,\qquad c=2\sqrt5$$

Step 2. The hyperbola

$ae=2\sqrt5$ with $e=5$:

$$a=\frac{2}{\sqrt5}$$$$b^2=c^2-a^2=20-\frac45=\frac{96}{5}$$

Step 3. Latus rectum

$$\frac{2b^2}{a}=\frac{192}{5}\cdot\frac{\sqrt5}{2}=\frac{96}{\sqrt5}$$

Answer: option (3)

Example 6 (JEE Main 2026, 23 January Shift 1). Let the domain of the function $f(x)=\log_3\log_5\log_7\left(9x-x^2-13\right)$ be the interval $(m,n)$. Let the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ have eccentricity $\frac n3$ and the length of the latus rectum $\frac{8m}{3}$. Then $b^2-a^2$ is equal to

▸ SHOW SOLUTION

Step 1. Domain

$\log_5\log_7(\cdot)>0$ needs $\log_7(\cdot)>1$:

$$9x-x^2-13>7$$$$(x-4)(x-5)<0$$

So $m=4$, $n=5$.

Step 2. The hyperbola

$e=\frac53$ gives $b^2=a^2\left(\frac{25}{9}-1\right)=\frac{16a^2}{9}$.

$$\frac{2b^2}{a}=\frac{32a}{9}=\frac{32}{3}$$

$a=3$, $b^2=16$.

Step 3. Required value

$$b^2-a^2=16-9=7$$

Answer: option (4)

Example 7 (JEE Main 2026, 4 April Shift 2). Let $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be a hyperbola such that the distance between its foci is $6$ and the distance between its directrices is $\frac83$. If the line $x=\alpha$ intersects $H$ at $A$ and $B$ such that the area of $\triangle AOB$ is $4\sqrt{15}$, where $O$ is the origin, then $\alpha^2$ equals

▸ SHOW SOLUTION

Step 1. Find a and e

$$2ae=6,\qquad \frac{2a}{e}=\frac83$$

Multiply and divide:

$$a^2=3\cdot\frac43=4,\qquad e^2=\frac{3}{4/3}=\frac94$$

$b^2=4\cdot\frac54=5$, so $H:\frac{x^2}{4}-\frac{y^2}{5}=1$.

Step 2. The chord AB

$$y^2=\frac{5\left(\alpha^2-4\right)}{4}$$$$AB=\sqrt{5\left(\alpha^2-4\right)}$$

Step 3. Area

$$\frac12\lvert\alpha\rvert\sqrt{5\left(\alpha^2-4\right)}=4\sqrt{15}$$$$\alpha^2\left(\alpha^2-4\right)=192$$$$\left(\alpha^2-16\right)\left(\alpha^2+12\right)=0$$

Answer: option (2)

Example 8 (JEE Main 2025, 2 April Shift 1). Let one focus of the hyperbola $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be at $\left(\sqrt{10},0\right)$ and the corresponding directrix be $x=\frac{9}{\sqrt{10}}$. If $e$ and $l$ respectively are the eccentricity and the length of the latus rectum of $H$, then $9\left(e^2+l\right)$ is equal to

▸ SHOW SOLUTION

Step 1. Focus and directrix

$$ae=\sqrt{10},\qquad \frac ae=\frac{9}{\sqrt{10}}$$$$a^2=9,\qquad e^2=\frac{10}{9}$$

Step 2. b and l

$$b^2=9\left(\frac{10}{9}-1\right)=1,\qquad l=\frac{2b^2}{a}=\frac23$$

Step 3. Required value

$$9\left(\frac{10}{9}+\frac23\right)=10+6=16$$

Answer: option (3)

Example 9 (JEE Main 2026, 6 April Shift 1). If the eccentricity $e$ of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, passing through $\left(6,4\sqrt3\right)$, satisfies $15\left(e^2+1\right)=34e$, then the length of the latus rectum of the hyperbola $\frac{x^2}{b^2}-\frac{y^2}{2\left(a^2+1\right)}=1$ is

▸ SHOW SOLUTION

Step 1. Eccentricity

$$15e^2-34e+15=0$$$$(3e-5)(5e-3)=0$$

For a hyperbola $e>1$, so $e=\frac53$ and $b^2=\frac{16a^2}{9}$.

Step 2. The point

$$\frac{36}{a^2}-\frac{48\cdot9}{16a^2}=1$$$$\frac{36-27}{a^2}=1,\qquad a^2=9,\ b^2=16$$

Step 3. Second hyperbola

$$\frac{x^2}{16}-\frac{y^2}{20}=1$$$$\text{latus rectum}=\frac{2\cdot20}{4}=10$$

Answer: option (1)

Example 10 (JEE Main 2026, 6 April Shift 2). The eccentricity of an ellipse $E$ with centre at the origin $O$ is $\frac{\sqrt3}{2}$ and its directrices are $x=\pm\frac{4\sqrt6}{3}$. Let $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be a hyperbola whose eccentricity equals the length of the semi-major axis of $E$, and whose length of latus rectum equals the length of the minor axis of $E$. Then the distance between the foci of $H$ is

▸ SHOW SOLUTION

Step 1. The ellipse

Write $A$, $B$ for the semi-axes of $E$ and $e_1=\frac{\sqrt3}{2}$.

$$\frac{A}{e_1}=\frac{4\sqrt6}{3},\qquad A=\frac{4\sqrt6}{3}\cdot\frac{\sqrt3}{2}=2\sqrt2$$$$B^2=8\left(1-\frac34\right)=2,\qquad 2B=2\sqrt2$$

Step 2. The hyperbola

$e=2\sqrt2$, so $b^2=a^2(8-1)=7a^2$.

$$\frac{2b^2}{a}=14a=2\sqrt2,\qquad a=\frac{\sqrt2}{7}$$

Step 3. Distance between foci

$$2ae=2\cdot\frac{\sqrt2}{7}\cdot2\sqrt2=\frac87$$

Answer: option (4)

Example 11 (JEE Main 2026, 28 January Shift 1). For some $\theta\in\left(0,\frac\pi2\right)$, let the eccentricity and length of the latus rectum of the hyperbola $x^2-y^2\sec^2\theta=8$ be $e_1$ and $\ell_1$, and those of the ellipse $x^2\sec^2\theta+y^2=6$ be $e_2$ and $\ell_2$. If $e_1^2=e_2^2\left(\sec^2\theta+1\right)$, then $\left(\frac{\ell_1\ell_2}{e_1e_2}\right)\tan^2\theta$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Standard forms

Write $k=\cos^2\theta$.

$$\frac{x^2}{8}-\frac{y^2}{8k}=1,\qquad \frac{x^2}{6k}+\frac{y^2}{6}=1$$

Step 2. The parts

Hyperbola: $e_1^2=1+k$, $\ell_1=\frac{2\cdot8k}{2\sqrt2}=4\sqrt2\,k$.

Ellipse (major axis along $y$, since $6>6k$): $e_2^2=1-k$, $\ell_2=\frac{2\cdot6k}{\sqrt6}=2\sqrt6\,k$.

Step 3. Use the condition

$$1+k=(1-k)\left(\frac1k+1\right)$$$$k+k^2=1-k^2$$$$(2k-1)(k+1)=0,\qquad k=\frac12$$

Step 4. Required value

$e_1e_2=\sqrt{\frac32\cdot\frac12}=\frac{\sqrt3}{2}$, $\ell_1\ell_2=8\sqrt{12}\cdot\frac14=4\sqrt3$, $\tan^2\theta=1$.

$$\frac{4\sqrt3}{\sqrt3/2}\cdot1=8$$

Answer: $8$

Example 12 (JEE Advanced 2018, Paper 2). Let $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, where $a>b>0$, be a hyperbola in the $xy$-plane whose conjugate axis $LM$ subtends an angle of $60^\circ$ at one of its vertices $N$. Let the area of the triangle $LMN$ be $4\sqrt3$.

▸ SHOW SOLUTION

Step 1. The angle

$L(0,b)$, $M(0,-b)$, $N(a,0)$. The angle $LNM=60^\circ$ is bisected by the $x$-axis:

$$\tan30^\circ=\frac ba,\qquad a=\sqrt3\,b$$

Step 2. The area

$$\frac12\cdot2b\cdot a=\sqrt3\,b^2=4\sqrt3$$

$b=2$, $a=2\sqrt3$.

Step 3. The parts

$$2b=4,\qquad e^2=1+\frac4{12}=\frac43$$$$e=\frac2{\sqrt3}$$$$2ae=2\cdot2\sqrt3\cdot\frac{2}{\sqrt3}=8,\qquad \frac{2b^2}{a}=\frac{4}{\sqrt3}$$

P→4, Q→3, R→1, S→2.

Answer: option (B)

When the ellipse and the hyperbola meet

Many questions give an ellipse and build a hyperbola from it: same foci (confocal), reciprocal eccentricities, or one passing through the foci or vertices of the other. Write both centres, both $c$ values, and match.

Example 13 (IIT JEE 2011, Paper 1). Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be reciprocal to that of the ellipse $x^2+4y^2=4$. If the hyperbola passes through a focus of the ellipse, then (one or more correct)

▸ SHOW SOLUTION

Step 1. The ellipse

$$\frac{x^2}{4}+y^2=1,\qquad e_1^2=1-\frac14=\frac34$$

Foci $\left(\pm\sqrt3,0\right)$.

Step 2. The hyperbola

$e^2=\frac43$, so $b^2=\frac{a^2}{3}$. It passes through $\left(\sqrt3,0\right)$, a point on its transverse axis, so $a^2=3$, $b^2=1$:

$$\frac{x^2}{3}-y^2=1\iff x^2-3y^2=3$$

Step 3. Focus

$$ae=\sqrt3\cdot\frac{2}{\sqrt3}=2$$

Foci $(\pm2,0)$. (A) and (C) are false.

Answer: (B) and (D)

Example 14 (JEE Main 2025, 7 April Shift 2). Let $e_1$ and $e_2$ be the eccentricities of the ellipse $\frac{x^2}{b^2}+\frac{y^2}{25}=1$ and the hyperbola $\frac{x^2}{16}-\frac{y^2}{b^2}=1$, respectively. If $b<5$ and $e_1e_2=1$, then the eccentricity of the ellipse having its axes along the coordinate axes and passing through all four foci (two of the ellipse and two of the hyperbola) is

▸ SHOW SOLUTION

Step 1. The two eccentricities

$$e_1^2=1-\frac{b^2}{25},\qquad e_2^2=1+\frac{b^2}{16}$$

Step 2. Use e1 e2 = 1

$$\left(25-b^2\right)\left(16+b^2\right)=400$$$$9b^2-b^4=0,\qquad b^2=9$$

Step 3. The four foci

Ellipse: $c^2=25-9=16$, foci $(0,\pm4)$. Hyperbola: $c^2=16+9=25$, foci $(\pm5,0)$.

Step 4. The new ellipse

$$\frac{x^2}{25}+\frac{y^2}{16}=1,\qquad e^2=1-\frac{16}{25}=\frac{9}{25}$$

Answer: option (2)

Example 15 (JEE Advanced 2026, Paper 2). Consider the ellipse $E$ given by $\frac{x^2}{18}+\frac{y^2}{12}=1$. Let $H$ be the hyperbola whose eccentricity is the reciprocal of the eccentricity of $E$ and whose foci are the same as that of $E$. Let $P$ and $Q$ be the points of intersection of $H$ and the parabola $\sqrt5\,y=x^2$ in the first quadrant. Let $d$ be the distance between $P$ and $Q$. If $a$ and $b$ are integers such that $d^2=a+b\sqrt5$, then the value of $a-b$ is ______.

▸ SHOW SOLUTION

Step 1. The ellipse

$$e_E^2=1-\frac{12}{18}=\frac13$$

Foci $\left(\pm\sqrt{18}\cdot\frac{1}{\sqrt3},0\right)=\left(\pm\sqrt6,0\right)$.

Step 2. The hyperbola

$e_H=\sqrt3$ and $Ae_H=\sqrt6$, so $A^2=2$, $B^2=2(3-1)=4$:

$$H:\ \frac{x^2}{2}-\frac{y^2}{4}=1$$

Step 3. Meet the parabola

Put $x^2=\sqrt5\,y$:

$$\frac{\sqrt5\,y}{2}-\frac{y^2}{4}=1$$$$y^2-2\sqrt5\,y+4=0,\qquad y=\sqrt5\pm1$$

Both are positive, and $x^2=5\pm\sqrt5>0$.

Step 4. Distance

$$(\Delta y)^2=4$$$$(\Delta x)^2=x_1^2+x_2^2-2x_1x_2=10-2\sqrt{20}$$$$d^2=14-4\sqrt5$$

$a=14$, $b=-4$, $a-b=18$.

Answer: $18$

Latus rectum and focal distances

For $P(x_1,y_1)$ on the hyperbola, the directrix $x=\frac ae$ gives

$$SP=e\left\lvert x_1-\frac ae\right\rvert=\lvert ex_1-a\rvert$$ $$S'P=\lvert ex_1+a\rvert$$

On the right branch the sum $SP+S'P=2ex_1$ and the product $SP\cdot S'P=e^2x_1^2-a^2$ depend on $x_1$; only the difference is constant.

Example 16 (JEE Main 2025, 4 April Shift 2). Let the sum of the focal distances of the point $P(4,3)$ on the hyperbola $H:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be $8\sqrt{\frac53}$. If for $H$, the length of the latus rectum is $l$ and the product of the focal distances of the point $P$ is $m$, then $9l^2+6m$ is equal to

▸ SHOW SOLUTION

Step 1. Sum of focal distances

$P$ is on the right branch:

$$(4e-a)+(4e+a)=8e=8\sqrt{\frac53}$$$$e^2=\frac53$$

Step 2. Find a and b

$b^2=\frac{2a^2}{3}$, and $P$ lies on $H$:

$$\frac{16}{a^2}-\frac{27}{2a^2}=1,\qquad a^2=\frac52,\ b^2=\frac53$$

Step 3. l and m

$$l^2=\frac{4b^4}{a^2}=\frac{4\cdot\frac{25}{9}}{\frac52}=\frac{40}{9}$$$$m=e^2\cdot16-a^2=\frac{80}{3}-\frac52=\frac{145}{6}$$

Step 4. Required value

$$9l^2+6m=40+145=185$$

Answer: option (3)

Example 17 (JEE Main 2026, 22 January Shift 2). Let $P\left(10,2\sqrt{15}\right)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, whose foci are $S$ and $S'$. If the length of its latus rectum is $8$, then the square of the area of $\triangle PSS'$ is equal to

▸ SHOW SOLUTION

Step 1. Find a and b

$\frac{2b^2}{a}=8$ gives $b^2=4a$:

$$\frac{100}{a^2}-\frac{60}{4a}=1$$$$a^2+15a-100=0,\qquad a=5,\ b^2=20$$

Step 2. Base SS'

$$a^2e^2=25+20=45$$$$SS'=2\sqrt{45}=6\sqrt5$$

Step 3. Area

$$\frac12\cdot6\sqrt5\cdot2\sqrt{15}=30\sqrt3$$$$\left(30\sqrt3\right)^2=2700$$

Answer: option (4)

Example 18 (JEE Main 2025, 7 April Shift 1). Consider the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having one of its foci at $P(-3,0)$. If the latus rectum through its other focus subtends a right angle at $P$ and $a^2b^2=\alpha\sqrt2-\beta$, $\alpha,\beta\in\mathbb N$, then $\alpha+\beta$ is ______.

▸ SHOW SOLUTION

Step 1. The right angle

The other focus is $S(3,0)$, the latus rectum ends are $L\left(3,\frac{b^2}{a}\right)$, $L'\left(3,-\frac{b^2}{a}\right)$. Triangle $LPL'$ is isosceles with the right angle at $P$, so

$$SL=PS,\qquad \frac{b^2}{a}=6$$

Step 2. Find a

$b^2=6a$ and $a^2+b^2=9$:

$$a^2+6a-9=0,\qquad a=3\left(\sqrt2-1\right)$$

Step 3. The product

$$a^2b^2=6a^3=162\left(\sqrt2-1\right)^3$$$$\left(\sqrt2-1\right)^3=5\sqrt2-7$$$$a^2b^2=810\sqrt2-1134$$

$\alpha=810$, $\beta=1134$.

Answer: $1944$

Auxiliary circle and the point θ

The circle on the transverse axis as diameter,

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

is the auxiliary circle. Take $Q(a\cos\theta,a\sin\theta)$ on it; the tangent to the circle at $Q$ meets the $x$-axis at $N(a\sec\theta,0)$. The perpendicular at $N$ meets the hyperbola at

$$P(a\sec\theta,\,b\tan\theta)$$

and $\theta$ is the eccentric angle of $P$. Since $\sec^2\theta-\tan^2\theta=1$, every such point lies on the curve.

The hyperbola and its auxiliary circle; a point Q on the circle at angle theta from the centre, the tangent at Q meeting the x-axis at N, and the point P on the hyperbola vertically above N

A second parametric form uses $\left(\frac a2\left(t+\frac1t\right),\frac b2\left(t-\frac1t\right)\right)$; it covers the right branch for $t>0$.

Example 19 (JEE Main 2026, 23 January Shift 2). Let $PQ$ be a chord of the hyperbola $\frac{x^2}{4}-\frac{y^2}{b^2}=1$, perpendicular to the $x$-axis, such that $OPQ$ is an equilateral triangle ($O$ being the centre of the hyperbola). If the eccentricity of the hyperbola is $\sqrt3$, then the area of the triangle $OPQ$ is

▸ SHOW SOLUTION

Step 1. The hyperbola

$$b^2=4(3-1)=8$$

Step 2. The point P

$P(2\sec\theta,2\sqrt2\tan\theta)$. In an equilateral triangle the line $OP$ makes $30^\circ$ with the axis:

$$\frac{2\sqrt2\tan\theta}{2\sec\theta}=\frac{1}{\sqrt3}$$$$\sin\theta=\frac{1}{\sqrt6},\qquad \sec^2\theta=\frac65$$

Step 3. Side and area

$x_P^2=4\sec^2\theta=\frac{24}{5}$ and $PQ=2y_P=\frac{2x_P}{\sqrt3}$:

$$\text{Area}=\frac{\sqrt3}{4}\cdot\frac{4x_P^2}{3}=\frac{x_P^2}{\sqrt3}$$$$\frac{24}{5\sqrt3}=\frac{8\sqrt3}{5}$$

Answer: option (2)

Conjugate hyperbola

The hyperbola whose transverse and conjugate axes are the conjugate and transverse axes of a given hyperbola is its conjugate hyperbola:

$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\quad\longleftrightarrow\quad\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$$ The hyperbola x squared over a squared minus y squared over b squared equals 1 with branches left and right, its conjugate hyperbola with branches above and below, the common asymptotes as dashed lines and the rectangle with corners at plus or minus a, plus or minus b

Both have the same asymptotes, and their four foci lie on the circle $x^2+y^2=a^2+b^2$. From the table,

$$\frac{1}{e_1^2}+\frac{1}{e_2^2}=\frac{a^2}{a^2+b^2}+\frac{b^2}{a^2+b^2}=1$$

Example 20 (JEE Main 2026, 28 January Shift 2). Let the ellipse $E:\frac{x^2}{144}+\frac{y^2}{169}=1$ and the hyperbola $H:\frac{x^2}{16}-\frac{y^2}{\lambda^2}=-1$ have the same foci. If $e$ and $L$ denote the eccentricity and length of the latus rectum of $H$, then $24(e+L)$ is

▸ SHOW SOLUTION

Step 1. Foci of the ellipse

The major axis is along $y$:

$$c^2=169-144=25,\qquad \text{foci }(0,\pm5)$$

Step 2. The hyperbola

$H$ is $\frac{y^2}{\lambda^2}-\frac{x^2}{16}=1$, transverse axis along $y$, foci $\left(0,\pm\sqrt{\lambda^2+16}\right)$:

$$\lambda^2+16=25,\qquad \lambda=3$$

Step 3. e and L

$$e=\frac53,\qquad L=\frac{2\cdot16}{3}=\frac{32}{3}$$$$24\left(\frac53+\frac{32}{3}\right)=24\cdot\frac{37}{3}=296$$

Answer: option (1)

Other forms of the hyperbola

Axes parallel to the coordinate axes

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

has centre $(h,k)$; all the parts of the standard table shift by $(h,k)$. The form

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

has transverse axis vertical, of length $2b$, with $a^2=b^2\left(e^2-1\right)$ and latus rectum $\frac{2a^2}{b}$.

A hyperbola with centre at (h,k) away from the origin, its transverse axis parallel to the x-axis drawn dashed, foci S and S' and vertices marked on it

Example 21 (IITianForum). The equation $9x^2-16y^2-18x-64y-199=0$ represents a hyperbola. Then (one or more correct)

▸ SHOW SOLUTION

Step 1. Complete the squares

$$9\left(x^2-2x\right)-16\left(y^2+4y\right)=199$$$$9(x-1)^2-16(y+2)^2=199+9-64$$$$9(x-1)^2-16(y+2)^2=144$$$$\frac{(x-1)^2}{16}-\frac{(y+2)^2}{9}=1$$

Step 2. The parts

Centre $(1,-2)$, $a=4$, $b=3$, transverse axis $8$.

$$e^2=1+\frac{9}{16}=\frac{25}{16},\qquad e=\frac54$$

Step 3. Foci

$ae=5$, so the foci are $(1\pm5,-2)$: $(6,-2)$ and $(-4,-2)$.

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

Example 22 (JEE Main 2026, 5 April Shift 2). Let the eccentricity $e$ of a hyperbola satisfy $6e^2-11e+3=0$. If the foci of the hyperbola are $(3,5)$ and $(3,-4)$, then the length of its latus rectum is

▸ SHOW SOLUTION

Step 1. Eccentricity

$$(2e-3)(3e-1)=0,\qquad e=\frac32$$

($e=\frac13<1$ is rejected.)

Step 2. The transverse axis

The foci lie on $x=3$, so the transverse axis is vertical, with

$$2be=9,\qquad b=3$$

where $b$ is now the semi-transverse axis.

Step 3. Latus rectum

$$a^2=b^2\left(e^2-1\right)=9\cdot\frac54=\frac{45}{4}$$$$\frac{2a^2}{b}=\frac{45/2}{3}=\frac{15}{2}$$

Answer: option (3)

Example 23 (JEE Main 2025, 3 April Shift 2). If the equation of the hyperbola with foci $(4,2)$ and $(8,2)$ is $3x^2-y^2-\alpha x+\beta y+\gamma=0$, then $\alpha+\beta+\gamma$ is equal to ______.

▸ SHOW SOLUTION

Step 1. Centre and c

Centre $(6,2)$, $ae=2$.

Step 2. Ratio from the given form

The coefficients of $x^2$ and $y^2$ give $\frac{b^2}{a^2}=3$:

$$a^2+3a^2=4,\qquad a^2=1,\ b^2=3$$

Step 3. Expand

$$3(x-6)^2-(y-2)^2=3$$$$3x^2-y^2-36x+4y+101=0$$

$\alpha=36$, $\beta=4$, $\gamma=101$.

Answer: $141$

The rectangular hyperbola

When $a=b$ the hyperbola is rectangular (or equilateral):

$$x^2-y^2=a^2,\qquad e=\sqrt2$$

Its asymptotes $y=\pm x$ are perpendicular. Turning the axes through $45^\circ$ turns $x^2-y^2=a^2$ into

$$xy=c^2,\qquad c^2=\frac{a^2}{2}$$

with the coordinate axes as asymptotes.

Two panels: the rectangular hyperbola x squared minus y squared equals a squared with the lines y equals x and y equals minus x dashed; the curve xy equals c squared in the first and third quadrants with the coordinate axes as asymptotes and the point (ct, c over t) marked

The section on $xy=c^2$ below gives its tangents, normals and chords.

Position of a point and a line meeting the hyperbola

Position of a point

For $P(x_1,y_1)$ write

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

From a point inside, no tangent can be drawn. From a point outside, two tangents can be drawn, except from points on the asymptotes (one tangent) and from the centre (none).

A line meeting the hyperbola

Put $y=mx+c$ in $b^2x^2-a^2y^2=a^2b^2$:

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

The discriminant divided by $4a^2b^2$ is $c^2-\left(a^2m^2-b^2\right)$.

A hyperbola with its asymptotes dashed and three lines: one of small slope crossing both branches, one parallel to an asymptote meeting the curve once, and one of large slope tangent to the right branch

Example 24 (JEE Main 2026, 22 January Shift 1). If the line $\alpha x+2y=1$, $\alpha\in\mathbb R$, does not meet the hyperbola $x^2-9y^2=9$, then a possible value of $\alpha$ is

▸ SHOW SOLUTION

Step 1. Substitute

$y=\frac{1-\alpha x}{2}$ in $4x^2-9\cdot4y^2=36$:

$$\left(4-9\alpha^2\right)x^2+18\alpha x-45=0$$

Step 2. No real root

When $9\alpha^2=4$ the equation is linear and has a root. Otherwise the discriminant must be negative:

$$324\alpha^2+180\left(4-9\alpha^2\right)<0$$$$\alpha^2>\frac59,\qquad \lvert\alpha\rvert>0.745$$

Only $0.8$ qualifies. (Check with the table: slope $\frac{\alpha}{2}=0.4>\frac13=\frac ba$.)

Answer: option (2)

Example 25 (IITianForum). From which of the following points can two tangents be drawn to $\frac{x^2}{9}-\frac{y^2}{16}=1$ that touch different branches? (one or more correct)

▸ SHOW SOLUTION

Step 1. Tangents through (h, k)

$y=mx+c$ with $c=k-mh$ touches when $c^2=9m^2-16$:

$$\left(h^2-9\right)m^2-2hkm+k^2+16=0$$

The point of contact of $y=mx+c$ is $\left(-\frac{9m}{c},-\frac{16}{c}\right)$, so the branch is decided by the sign of $-\frac mc$.

Step 2. Point (2, 5)

$$-5m^2-20m+41=0$$

The roots are $m\approx1.49$ and $m\approx-5.49$. Here $c=5-2m>0$ for both; the contact abscissa $-\frac{9m}{c}$ is negative for the first and positive for the second. Different branches.

Step 3. Point (3, 4)

$(3,4)$ lies on the asymptote $y=\frac43x$: $-24m+32=0$ gives $m=\frac43$, $c=0$, which is the asymptote itself. Only one tangent exists (the vertical line $x=3$).

Step 4. Point (−3, 2)

$S_1=1-\frac{4}{16}-1<0$, so the point is outside. Here $h^2-9=0$: $12m+20=0$, $m=-\frac53$, $c=2-5=-3$, contact $x=-\frac{9m}{c}=-5$. The other tangent is the vertical line $x=-3$, touching at $(-3,0)$. Both are on the left branch.

Step 5. Point (0, 1)

$$-9m^2+17=0,\qquad m=\pm\frac{\sqrt{17}}{3},\ c=1$$

Contact $x=-9m$: opposite signs, different branches.

Step 6. The rule

Tangents touch different branches exactly when the point lies between the asymptotes in the region containing the conjugate axis: $\lvert k\rvert>\frac ba\lvert h\rvert$. For $(2,5)$: $5>\frac83$; for $(0,1)$: $1>0$.

Answer: (A) and (D)

A hyperbola with its asymptotes dashed, a point T on the conjugate axis between the asymptotes, and the two tangents from T, one touching the right branch and one touching the left branch

Example 26 (IIT JEE 2010, Paper 1). The circle $x^2+y^2-8x=0$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$. The equation of the circle with $AB$ as its diameter is

▸ SHOW SOLUTION

Step 1. Eliminate y

$y^2=\frac{4x^2}{9}-4$ in the circle:

$$13x^2-72x-36=0$$$$(x-6)(13x+6)=0$$

Step 2. Reject the extra root

At $x=-\frac{6}{13}$, $\lvert x\rvert<3$, so $y^2<0$. At $x=6$, $y^2=12$:

$$A\left(6,2\sqrt3\right),\qquad B\left(6,-2\sqrt3\right)$$

Step 3. Circle on AB

Centre $(6,0)$, radius $2\sqrt3$:

$$(x-6)^2+y^2=12$$

Answer: option (A)

Tangents

The three forms of the tangent

The slope form touches at

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

For a line $lx+my=n$ the tangency condition is $a^2l^2-b^2m^2=n^2$.

The tangents at $\theta_1$ and $\theta_2$ meet at

$$x=\frac{a\cos\frac{\theta_1-\theta_2}{2}}{\cos\frac{\theta_1+\theta_2}{2}},\qquad y=b\tan\frac{\theta_1+\theta_2}{2}$$ The hyperbola with a point P on the right branch, the tangent at P bisecting the angle between PS and PS', the normal at P meeting the x-axis at G

Example 27 (IIT JEE 2010). The line $2x+y=1$ is tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If this line passes through the point of intersection of the nearest directrix and the $x$-axis, then find the eccentricity of the hyperbola.

▸ SHOW SOLUTION

Step 1. The directrix point

The line cuts the $x$-axis at $\left(\frac12,0\right)$, which is $\left(\frac ae,0\right)$:

$$a=\frac e2$$

Step 2. Tangency

$y=-2x+1$ with $c^2=a^2m^2-b^2$:

$$1=4a^2-a^2\left(e^2-1\right)=a^2\left(5-e^2\right)$$

Step 3. Solve

$$\frac{e^2}{4}\left(5-e^2\right)=1$$$$e^4-5e^2+4=0,\qquad e^2=4\ \text{or}\ 1$$

$e>1$, so $e=2$.

Answer: $e=2$

Example 28 (JEE Main 2017). A hyperbola passes through the point $P\left(\sqrt2,\sqrt3\right)$ and has foci at $(\pm2,0)$. Then the tangent to this hyperbola at $P$ also passes through the point

▸ SHOW SOLUTION

Step 1. The hyperbola

$a^2+b^2=4$ and $\frac{2}{a^2}-\frac{3}{b^2}=1$. With $b^2=4-a^2$:

$$a^4-9a^2+8=0,\qquad a^2=1$$

($a^2=8>4$ is rejected.) So $b^2=3$.

Step 2. Tangent at P

$$\sqrt2\,x-\frac{\sqrt3\,y}{3}=1$$

Step 3. Test the options

For $\left(2\sqrt2,3\sqrt3\right)$: $4-3=1$.

Answer: option (2)

Example 29 (IIT JEE 2012, Paper 1). Tangents are drawn to the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$, parallel to the straight line $2x-y=1$. The points of contact of the tangents on the hyperbola are (one or more correct)

▸ SHOW SOLUTION

Step 1. Tangents of slope 2

$$c^2=9\cdot4-4=32,\qquad c=\pm4\sqrt2$$

Step 2. Points of contact

$$\left(-\frac{a^2m}{c},-\frac{b^2}{c}\right)=\left(-\frac{18}{c},-\frac4c\right)$$

$c=-4\sqrt2$: $\left(\frac{9}{2\sqrt2},\frac{1}{\sqrt2}\right)$. $c=4\sqrt2$: $\left(-\frac{9}{2\sqrt2},-\frac{1}{\sqrt2}\right)$.

Answer: (A) and (B)

Example 30 (JEE Advanced 2017, Paper 1). If $2x-y+1=0$ is a tangent to the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{16}=1$, then which of the following CANNOT be sides of a right angled triangle? (one or more correct)

▸ SHOW SOLUTION

Step 1. Find a

$y=2x+1$:

$$1=4a^2-16,\qquad a^2=\frac{17}{4},\ 2a=\sqrt{17}$$

Step 2. Test each triple

Triple	Largest$^2$	Sum of other squares	Right angled?
$a,4,1$	$16$	$\frac{17}{4}+1$	no
$2a,4,1$	$17$	$16+1$	yes
$a,4,2$	$16$	$\frac{17}{4}+4$	no
$2a,8,1$	$64$	$17+1$	no

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

Example 31 (JEE Main 2021, 18 March Shift 2). Consider a hyperbola $H:x^2-2y^2=4$. Let the tangent at a point $P\left(4,\sqrt6\right)$ meet the $x$-axis at $Q$ and the latus rectum at $R\left(x_1,y_1\right)$, $x_1>0$. If $F$ is a focus of $H$ which is nearer to the point $P$, then the area of $\triangle QFR$ is equal to

▸ SHOW SOLUTION

Step 1. The hyperbola

$$\frac{x^2}{4}-\frac{y^2}{2}=1,\qquad e^2=\frac32,\ ae=\sqrt6$$

$F\left(\sqrt6,0\right)$.

Step 2. Tangent at P

$$x-\frac{\sqrt6}{2}y=1$$

$Q(1,0)$. At $x=\sqrt6$: $y_1=\frac{2\left(\sqrt6-1\right)}{\sqrt6}$.

Step 3. Area

Right angle at $F$:

$$\frac12\left(\sqrt6-1\right)\cdot\frac{2\left(\sqrt6-1\right)}{\sqrt6}=\frac{7-2\sqrt6}{\sqrt6}$$$$=\frac{7}{\sqrt6}-2$$

Answer: option (3)

Example 32 (IITianForum). The point of the hyperbola $\frac{x^2}{12}-\frac{y^2}{3}=1$ nearest to the line $x-y+1=0$ is

▸ SHOW SOLUTION

Step 1. Does the line meet the curve?

$m=1>\frac ba=\frac12$, and $c^2=1<a^2m^2-b^2=9$: no meeting point.

Step 2. Tangents parallel to the line

$$y=x\pm3$$

Contact of $y=x+c$: $\left(-\frac{12}{c},-\frac{3}{c}\right)$. $c=3$: $(-4,-1)$; $c=-3$: $(4,1)$.

Step 3. Nearer tangent

Distance of $y=x+1$ from $y=x+3$ is $\sqrt2$; from $y=x-3$ it is $2\sqrt2$. The nearest point is $(-4,-1)$, at distance $\sqrt2$.

Answer: option (B)

Pair of tangents and chord of contact

From an outside point $P(x_1,y_1)$, with

$$T=\frac{xx_1}{a^2}-\frac{yy_1}{b^2}-1$$ A point T outside the hyperbola, the two tangents from T touching the right branch at P and Q, and the chord of contact PQ

If the pole of a line lies on a second line, the pole of the second lies on the first (conjugate lines). If the poles of a family of lines lie on a fixed line, all the polars pass through a fixed point.

Director circle

Tangents $y=mx\pm\sqrt{a^2m^2-b^2}$ through $(h,k)$:

$$\left(h^2-a^2\right)m^2-2hkm+k^2+b^2=0$$

Perpendicular tangents need $m_1m_2=-1$, that is $k^2+b^2=a^2-h^2$:

$$x^2+y^2=a^2-b^2$$ The hyperbola with a greater than b, its director circle of radius root of a squared minus b squared, and a point on the circle with two perpendicular tangents to the hyperbola

Example 33 (IITianForum). Find the director circle of $\frac{x^2}{16}-\frac{y^2}{7}=1$. Verify that the tangents from $(0,3)$ are perpendicular and find them. What is the director circle of $\frac{x^2}{9}-\frac{y^2}{16}=1$?

▸ SHOW SOLUTION

Step 1. Director circle

$$x^2+y^2=16-7=9$$

Step 2. Tangents from (0, 3)

$(0,3)$ lies on $x^2+y^2=9$. With $c=3$:

$$9=16m^2-7,\qquad m=\pm1$$

The slopes multiply to $-1$. Tangents $y=x+3$ and $y=-x+3$.

Step 3. The second hyperbola

$a^2-b^2=9-16<0$: no point of the plane sends perpendicular tangents to it.

Answer: $x^2+y^2=9$; $y=\pm x+3$; none for $\frac{x^2}{9}-\frac{y^2}{16}=1$

Common tangents

Write the slope form for each curve and equate the intercepts.

Example 34 (IIT JEE 2010, Paper 1). The circle $x^2+y^2-8x=0$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ intersect at the points $A$ and $B$. The equation of a common tangent with positive slope to the circle as well as to the hyperbola is

▸ SHOW SOLUTION

Step 1. Two conditions

$y=mx+c$ touches the hyperbola: $c^2=9m^2-4$. It touches the circle with centre $(4,0)$, radius $4$:

$$(4m+c)^2=16\left(1+m^2\right)$$$$8mc+c^2=16$$

Step 2. Eliminate c

$c^2=9m^2-4$ gives $8mc=20-9m^2$; squaring and using $c^2$ again:

$$495m^4+104m^2-400=0$$$$m^2=\frac45,\qquad m=\frac{2}{\sqrt5}$$

Step 3. The intercept

$$c=\frac{20-\frac{36}{5}}{\frac{16}{\sqrt5}}=\frac{4}{\sqrt5}$$$$\sqrt5\,y=2x+4$$

Answer: option (B)

Example 35 (IITianForum). The common tangents of $x^2-y^2=8$ and $xy=3$ are (one or more correct)

▸ SHOW SOLUTION

Step 1. Tangency to xy = 3

$y=mx+c$ in $xy=3$:

$$mx^2+cx-3=0$$

Equal roots: $c^2+12m=0$.

Step 2. Tangency to x² − y² = 8

$$c^2=8m^2-8$$

Step 3. Solve

$$8m^2+12m-8=0$$$$(2m-1)(m+2)=0$$

$c^2=-12m\ge0$ needs $m<0$: $m=-2$, $c^2=24$.

$$y=-2x\pm2\sqrt6$$

Answer: (A) and (C)

Normals

The forms of the normal

The normal at $(x_1,y_1)$ meets the axes at

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

Co-normal points

Through a given point, at most four normals can be drawn; their feet are co-normal points. If $\theta_1,\theta_2,\theta_3,\theta_4$ are their eccentric angles:

$$\theta_1+\theta_2+\theta_3+\theta_4=(2n+1)\pi$$ $$\sum\sin\left(\theta_i+\theta_j\right)=0$$ $$\sum\cos\left(\theta_i+\theta_j\right)=0$$

and, for the abscissae of the four feet,

$$\left(\sum x_i\right)\left(\sum\frac{1}{x_i}\right)=4$$

Example 36 (IIT JEE 2011, Paper 2). Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. If the normal at the point $P$ intersects the $x$-axis at $(9,0)$, then the eccentricity of the hyperbola is

▸ SHOW SOLUTION

Step 1. Where the normal meets the axis

$$G\left(e^2x_1,0\right)=(9,0)$$

Step 2. Solve

$$6e^2=9,\qquad e=\sqrt{\frac32}$$

Answer: option (B)

Example 37 (JEE Main 2021, 26 August Shift 2). The point $P\left(-2\sqrt6,\sqrt3\right)$ lies on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ having eccentricity $\frac{\sqrt5}{2}$. If the tangent and normal at $P$ to the hyperbola intersect its conjugate axis at the points $Q$ and $R$ respectively, then $QR$ is equal to

▸ SHOW SOLUTION

Step 1. The hyperbola

$b^2=\frac{a^2}{4}$:

$$\frac{24}{a^2}-\frac{12}{a^2}=1,\qquad a^2=12,\ b^2=3$$

Step 2. Tangent meets x = 0

$$\frac{-2\sqrt6\,x}{12}-\frac{\sqrt3\,y}{3}=1$$

$x=0$: $y=-\sqrt3$.

Step 3. Normal meets x = 0

$$\frac{12x}{-2\sqrt6}+\frac{3y}{\sqrt3}=15$$

$x=0$: $y=5\sqrt3$.

$$QR=5\sqrt3+\sqrt3=6\sqrt3$$

Answer: option (3)

Example 38 (JEE Advanced 2020, Paper 2). Let $a$ and $b$ be positive real numbers such that $a>1$ and $b<a$. Let $P$ be a point in the first quadrant that lies on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$. Suppose the tangent to the hyperbola at $P$ passes through the point $(1,0)$, and suppose the normal to the hyperbola at $P$ cuts off equal intercepts on the coordinate axes. Let $\Delta$ denote the area of the triangle formed by the tangent at $P$, the normal at $P$ and the $x$-axis. If $e$ denotes the eccentricity of the hyperbola, then which of the following statements is/are TRUE?

▸ SHOW SOLUTION

Step 1. Tangent through (1, 0)

$\frac{x_1}{a^2}=1$, so $x_1=a^2$.

Step 2. Normal with equal intercepts

The normal has slope $-1$. Its slope is $-\frac{a^2y_1}{b^2x_1}$:

$$\frac{a^2y_1}{b^2a^2}=1,\qquad y_1=b^2$$

Step 3. P on the curve

$$\frac{a^4}{a^2}-\frac{b^4}{b^2}=1,\qquad a^2-b^2=1$$$$e^2=1+\frac{a^2-1}{a^2}=2-\frac{1}{a^2}$$

So $1<e^2<2$.

Step 4. The triangle

Tangent meets the axis at $(1,0)$; the normal $x+y=a^2+b^2$ at $\left(a^2+b^2,0\right)$. Height $y_1=b^2$:

$$\Delta=\frac12\left(a^2+b^2-1\right)b^2$$$$=\frac12\cdot2b^2\cdot b^2=b^4$$

Answer: (A) and (D)

Example 39 (JEE Advanced 2015, Paper 2). Consider the hyperbola $H:x^2-y^2=1$ and a circle $S$ with centre $N\left(x_2,0\right)$. Suppose that $H$ and $S$ touch each other at a point $P\left(x_1,y_1\right)$ with $x_1>1$ and $y_1>0$. The common tangent to $H$ and $S$ at $P$ intersects the $x$-axis at the point $M$. If $(l,m)$ is the centroid of the triangle $PMN$, then the correct expression(s) is(are)

▸ SHOW SOLUTION

Step 1. N is where the normal meets the axis

The radius $NP$ is along the normal at $P$. With $a=b=1$, $e^2=2$:

$$N=\left(2x_1,0\right)$$

Step 2. M from the tangent

$xx_1-yy_1=1$ at $y=0$: $M\left(\frac{1}{x_1},0\right)$.

Step 3. Centroid

$$l=\frac{x_1+\frac{1}{x_1}+2x_1}{3}=x_1+\frac{1}{3x_1}$$$$m=\frac{y_1}{3}=\frac{\sqrt{x_1^2-1}}{3}$$

Step 4. Derivatives

$$\frac{dl}{dx_1}=1-\frac{1}{3x_1^2},\qquad \frac{dm}{dx_1}=\frac{x_1}{3\sqrt{x_1^2-1}}$$

and $\frac{dm}{dy_1}=\frac13$.

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

Chords of the hyperbola

Chord joining two points

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

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

Focal chords

Putting $S(ae,0)$ in the chord:

$$e=\frac{\cos\frac{\theta+\phi}{2}}{\cos\frac{\theta-\phi}{2}}$$

With $\tan\frac\theta2\tan\frac\phi2=p$, this ratio is $\frac{1-p}{1+p}$:

For a focal chord $PQ$ through $S$, $\frac{1}{SP}+\frac{1}{SQ}=\frac{2a}{b^2}$ when $P$, $Q$ are on the same branch (the semi-latus rectum is their harmonic mean).

A focal chord PQ of the right branch passing through the focus S, with the other focus S' marked

Example 40 (IITianForum). The points $(2\sec\theta,4\sqrt2\tan\theta)$ and $(2\sec\phi,4\sqrt2\tan\phi)$ are the ends of a focal chord of $\frac{x^2}{4}-\frac{y^2}{32}=1$. Then $\tan\frac\theta2\tan\frac\phi2$ can be (one or more correct)

▸ SHOW SOLUTION

Step 1. Eccentricity

$$e^2=1+\frac{32}{4}=9,\qquad e=3$$

Step 2. Through S(6, 0)

$$\frac{1-e}{1+e}=\frac{-2}{4}=-\frac12$$

Step 3. Through S'(−6, 0)

$$\frac{1+e}{1-e}=\frac{4}{-2}=-2$$

Answer: (A) and (B)

Chord with a given middle point

The chord with middle point $(x_1,y_1)$ is

$$T=S_1$$ $$\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=\frac{x_1^2}{a^2}-\frac{y_1^2}{b^2}$$

Its slope is $\frac{b^2x_1}{a^2y_1}$. For $xy=c^2$ the same rule gives $\frac{x}{x_1}+\frac{y}{y_1}=2$.

Diameters

The middle points of all chords of slope $m$ lie on

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

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

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

Of two conjugate diameters, exactly one meets the hyperbola; the other meets the conjugate hyperbola.

Example 41 (JEE Main 2021, 26 August Shift 2). The locus of the mid points of the chords of the hyperbola $x^2-y^2=4$, which touch the parabola $y^2=8x$, is

▸ SHOW SOLUTION

Step 1. Chord with middle point (h, k)

$T=S_1$:

$$hx-ky=h^2-k^2$$$$y=\frac hk\,x-\frac{h^2-k^2}{k}$$

Step 2. Touches y² = 8x

$y=mx+c$ touches when $c=\frac2m$:

$$-\frac{h^2-k^2}{k}=\frac{2k}{h}$$$$k^2h-h^3=2k^2$$$$k^2(h-2)=h^3$$

Answer: option (3)

Example 42 (JEE Main 2018). Tangents are drawn to the hyperbola $4x^2-y^2=36$ at the points $P$ and $Q$. If these tangents intersect at the point $T(0,3)$, then the area (in sq. units) of $\triangle PTQ$ is

▸ SHOW SOLUTION

Step 1. Chord of contact

$T=0$ for $4x^2-y^2=36$ from $(0,3)$:

$$4\cdot0\cdot x-3y=36,\qquad y=-12$$

Step 2. P and Q

$$4x^2=36+144,\qquad x=\pm3\sqrt5$$

$PQ=6\sqrt5$.

Step 3. Area

Height from $T$ to $y=-12$ is $15$:

$$\frac12\cdot6\sqrt5\cdot15=45\sqrt5$$

Answer: option (3)

Example 43 (IITianForum). From each point of the line $x+y=4$, tangents are drawn to $\frac{x^2}{8}-\frac{y^2}{4}=1$. All the chords of contact pass through a fixed point $(\lambda,\mu)$. Then $\frac\lambda\mu$ is

▸ SHOW SOLUTION

Step 1. Chord of contact from (h, 4 − h)

$$\frac{hx}{8}-\frac{(4-h)y}{4}=1$$

Step 2. Separate h

$$h\left(\frac x8+\frac y4\right)-(y+1)=0$$

Step 3. Fixed point

$$y=-1,\qquad \frac x8=\frac14,\ x=2$$

$(2,-1)$ is the pole of $x+y=4$.

Answer: option (A)

Asymptotes

An asymptote is a line that the curve approaches as the point runs to infinity; it is the limit of the tangent. For the standard hyperbola,

$$\frac xa\pm\frac yb=0$$

Their joint equation is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=0$.

Hyperbola, asymptotes, conjugate

If $H=0$ is a hyperbola, $A=0$ its pair of asymptotes and $H'=0$ its conjugate, then $A$, $H$, $H'$ differ only in the constant term, with

$$H+H'=2A$$

To find the asymptotes of a general hyperbola, change only the constant so that the equation becomes a pair of lines ($\Delta=0$).

Angle between the asymptotes

The asymptotes make angles $\pm\tan^{-1}\frac ba$ with the transverse axis. The angle containing the hyperbola is

$$2\tan^{-1}\frac ba=2\sec^{-1}e$$

Standard results

The hyperbola with its asymptotes through C, the tangent at P meeting the asymptotes at L and G, P the midpoint of LG, and the triangle CLG shaded

The reason for the area: tangent at $\theta$ meets $y=\frac bax$ at $L\left(\frac{a}{\sec\theta-\tan\theta},\ldots\right)$ and $y=-\frac bax$ at $G\left(\frac{a}{\sec\theta+\tan\theta},\ldots\right)$; the product of these abscissae is $a^2$, and the area is $\frac ba\cdot x_Lx_G=ab$.

Example 44 (IITianForum). The asymptotes of a hyperbola are given by $2x^2+5xy+2y^2+3x+3y+\mu=0$. Then the eccentricity of the hyperbola is/are (one or more correct)

▸ SHOW SOLUTION

Step 1. The pair of lines

$$2x^2+5xy+2y^2=(2x+y)(x+2y)$$

so the asymptotes have slopes $-2$ and $-\frac12$. ($\Delta=0$ gives $\mu=1$; not needed for $e$.)

Step 2. Angle between them

$$\tan\psi=\left\lvert\frac{-2+\frac12}{1+1}\right\rvert=\frac34$$

Step 3. The angle containing the curve

The curve lies in the angle $\psi$ or in $\pi-\psi$. If $2\alpha=\psi$:

$$\frac{2t}{1-t^2}=\frac34,\qquad t=\tan\alpha=\frac13$$$$e^2=1+\frac19=\frac{10}{9}$$

If $2\alpha=\pi-\psi$: $\tan\alpha=3$, $e^2=10$.

The given data cannot separate the two cases.

Answer: (A) and (B)

Example 45 (IITianForum). A hyperbola has asymptotes $2x-y=0$ and $2x+y-4=0$ and passes through $P(3,2)$. Then (one or more correct)

▸ SHOW SOLUTION

Step 1. The hyperbola

$H=A+\text{constant}$:

$$(2x-y)(2x+y-4)=k$$

At $(3,2)$: $k=4\cdot4=16$.

Step 2. Standard form

$$4x^2-y^2-8x+4y=16$$$$4(x-1)^2-(y-2)^2=16$$$$\frac{(x-1)^2}{4}-\frac{(y-2)^2}{16}=1$$

Step 3. The parts

Transverse axis $y=2$, length $4$; $e^2=1+4=5$.

Step 4. Conjugate

$H'=2A-H$:

$$(2x-y)(2x+y-4)=-16$$$$4x^2-y^2-8x+4y+16=0$$

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

Example 46 (IITianForum). Let $H:\frac{(x+2)^2}{16}-\frac{(y-1)^2}{9}=1$ with centre $C$. From a point $Q$ on an asymptote, the line perpendicular to the transverse axis meets $H$ at $P$ and $P'$. The tangent at $P$ meets the asymptotes at $L$ and $G$. Then (one or more correct)

▸ SHOW SOLUTION

Step 1. Shift the centre

With $X=x+2$, $Y=y-1$: $\frac{X^2}{16}-\frac{Y^2}{9}=1$, asymptote $Y=\frac34X$.

Step 2. The vertical line X = u

$Q$ has $Y=\frac34u$; $P$, $P'$ have $Y=\pm3\sqrt{\frac{u^2}{16}-1}$:

$$PQ\cdot QP'=\frac{9u^2}{16}-9\left(\frac{u^2}{16}-1\right)=9$$

This is $b^2$.

Step 3. Triangle CLG

$$\text{Area}=ab=4\cdot3=12$$

Answer: (A) and (C)

The rectangular hyperbola xy = c²

For the point $t$, that is $\left(ct,\frac ct\right)$:

The reason for the orthocentre: the side $t_2t_3$ has slope $-\frac{1}{t_2t_3}$, so the altitude from $t_1$ has slope $t_2t_3$; the point $\left(-\frac{c}{t_1t_2t_3},-ct_1t_2t_3\right)$ satisfies all three altitude equations.

The curve xy equals c squared, a point P with its normal meeting the other branch again at Q

Example 47 (JEE Main 2026, 2 April Shift 2). Let $O$ be the origin, and $P$ and $Q$ be two points on the rectangular hyperbola $xy=12$ such that the mid point of the line segment $PQ$ is $\left(\frac12,-\frac12\right)$. Then the area of the triangle $OPQ$ equals

▸ SHOW SOLUTION

Step 1. Chord with middle point

$$\frac{x}{1/2}+\frac{y}{-1/2}=2,\qquad x-y=1$$

Step 2. Meet the curve

$$(y+1)y=12,\qquad y=3\ \text{or}\ -4$$

$P(4,3)$, $Q(-3,-4)$.

Step 3. Area

$$\frac12\left\lvert4(-4)-(-3)(3)\right\rvert=\frac72$$

Answer: option (C)

Example 48 (JEE Main 2019, 11 January Shift 1). Equation of a common tangent to the parabola $y^2=4x$ and the hyperbola $xy=2$ is

▸ SHOW SOLUTION

Step 1. Tangent to the parabola

$$y=mx+\frac1m$$

Step 2. Tangent to xy = 2

$mx^2+\frac xm-2=0$ needs equal roots:

$$\frac{1}{m^2}+8m=0,\qquad m^3=-\frac18,\ m=-\frac12$$

Step 3. The line

$$y=-\frac x2-2,\qquad x+2y+4=0$$

Answer: option (3)

Example 49 (IITianForum). The angle at which the curves $xy=4$ and $x^2-y^2=6$ cut each other is

▸ SHOW SOLUTION

Step 1. Slopes at a common point

$$xy=4:\quad m_1=-\frac yx$$$$x^2-y^2=6:\quad m_2=\frac xy$$

Step 2. Product

$$m_1m_2=-1$$

The curves cut at right angles at every common point, for any values of the constants.

Answer: option (D)

Example 50 (IITianForum). The normal to $xy=4$ at $P(4,1)$ meets the curve again at $Q$. Find $Q$ and $PQ$.

▸ SHOW SOLUTION

Step 1. Parameter of P

$c=2$, $P=\left(2t,\frac2t\right)$ with $t_1=2$.

Step 2. Second end

$$t_1^3t_2=-1,\qquad t_2=-\frac18$$$$Q\left(-\frac14,-16\right)$$

Step 3. Check

The normal at $P$ has slope $t_1^2=4$: $y-1=4(x-4)$. At $x=-\frac14$: $y=1-17=-16$.

Step 4. Length

$$PQ=\sqrt{\left(\frac{17}{4}\right)^2+17^2}=\frac{17\sqrt{17}}{4}$$

Answer: $Q\left(-\frac14,-16\right)$, $PQ=\frac{17\sqrt{17}}{4}$

Example 51 (IITianForum). Let $P_1(1,1)$, $P_2\left(3,\frac13\right)$, $P_3\left(-2,-\frac12\right)$, and for $n\ge4$ let $P_n$ be the orthocentre of the triangle $P_{n-1}P_{n-2}P_{n-3}$. Find $P_4$ and the abscissa of $P_{2026}$.

▸ SHOW SOLUTION

Step 1. The points are on xy = 1

$c=1$, with $t_1=1$, $t_2=3$, $t_3=-2$.

Step 2. Orthocentre rule

The orthocentre of $t_i,t_j,t_k$ is the point with parameter $-\frac{1}{t_it_jt_k}$:

$$t_4=-\frac{1}{1\cdot3\cdot(-2)}=\frac16,\qquad P_4\left(\frac16,6\right)$$

Step 3. The sequence repeats

$$t_5=-\frac{1}{3\cdot(-2)\cdot\frac16}=1=t_1$$

In the same way $t_6=t_2$, $t_7=t_3$: the period is $4$.

Step 4. P2026

$2026=4\cdot506+2$, so $P_{2026}=P_2$.

Answer: $P_4\left(\frac16,6\right)$; abscissa of $P_{2026}$ is $3$

Reflection property and standard focal results

Reflection property

The tangent at $P$ bisects the interior angle $SPS'$; the normal bisects the exterior angle. So a ray of light aimed at one focus is reflected by the hyperbola (mirror on the outside) towards the other focus; a ray from one focus, reflected on the inside, goes out as if it came from the other focus.

The right branch of a hyperbola with foci S and S', a ray travelling towards S that meets the outside of the branch at P and is reflected towards S', and the tangent at P

An ellipse and a hyperbola with the same foci cut at right angles, because at a common point the tangent to one bisects the interior angle $SPS'$ and the tangent to the other bisects the exterior angle.

More standard results

Example 52 (IITianForum). A ray of light travelling along the line $x=4$ strikes the hyperbola $\frac{x^2}{4}-\frac{y^2}{12}=1$ at $A(4,6)$ and is reflected. The reflected ray lies along

▸ SHOW SOLUTION

Step 1. The foci

$$a^2e^2=4+12=16$$$$S(4,0),\quad S'(-4,0)$$

Step 2. The incident ray

$x=4$ passes through $S$: the ray runs along $SA$.

Step 3. Reflection property

The tangent at $A$ bisects angle $SAS'$, so lines $AS$ and $AS'$ are mirror images in the tangent. The reflected ray lies along $AS'$:

$$\text{slope}=\frac{6-0}{4+4}=\frac34$$$$y-6=\frac34(x-4),\qquad 3x-4y+12=0$$

Answer: option (A)

Example 53 (IITianForum). A circle is drawn with centre at a focus of $\frac{x^2}{16}-\frac{y^2}{20}=1$ so that it touches the hyperbola and no part of it lies outside the branch. Find its radius.

▸ SHOW SOLUTION

Step 1. The focus

$$a^2e^2=16+20=36,\qquad S(6,0),\ e=\frac32$$

Step 2. Distance from S to the branch

For $P(x_1,y_1)$ on the right branch:

$$SP=ex_1-a\ge ea-a=6-4=2$$

with equality at the vertex $(4,0)$.

Step 3. Radius

The largest circle about $S$ inside the branch touches it at the vertex.

Answer: $2$

Example 54 (IITianForum). A hyperbola $H$ with axes along the coordinate axes cuts the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ at right angles, and its eccentricity is $\frac52$ times that of the ellipse. If $2a$, $2b$ are the transverse and conjugate axes of $H$ and $\ell$ its latus rectum, find $a+b^2+\ell$.

▸ SHOW SOLUTION

Step 1. Right angles means same foci

At a common point $(x,y)$ the slopes are $-\frac{9x}{25y}$ and $\frac{b^2x}{a^2y}$; their product is $-1$ exactly when $25-9=a^2+b^2$. So $H$ is confocal with the ellipse.

Step 2. The ellipse

$$e_1=\frac45,\qquad \text{foci }(\pm4,0)$$

Step 3. The hyperbola

$e=\frac52\cdot\frac45=2$, $ae=4$:

$$a=2,\qquad b^2=16-4=12$$$$\ell=\frac{2\cdot12}{2}=12$$

Step 4. Required value

$$a+b^2+\ell=2+12+12=26$$

Answer: $26$

Example 55 (JEE Advanced 2022, Paper 2). Consider the hyperbola $\frac{x^2}{100}-\frac{y^2}{64}=1$ with foci at $S$ and $S_1$, where $S$ lies on the positive $x$-axis. Let $P$ be a point on the hyperbola, in the first quadrant. Let $\angle SPS_1=\alpha$, with $\alpha<\frac\pi2$. The straight line passing through the point $S$ and having the same slope as that of the tangent at $P$ to the hyperbola, intersects the straight line $S_1P$ at $P_1$. Let $\delta$ be the distance of $P$ from the straight line $SP_1$, and $\beta=S_1P$. Then the greatest integer less than or equal to $\frac{\beta\delta}{9}\sin\frac\alpha2$ is ______.

▸ SHOW SOLUTION

Step 1. The angle between PS and the tangent

The tangent at $P$ bisects angle $SPS_1$, so it makes angle $\frac\alpha2$ with $PS$.

Step 2. Find δ

$SP_1$ is parallel to the tangent and passes through $S$:

$$\delta=SP\sin\frac\alpha2$$

Step 3. Focal triangle

$$\frac{\beta\delta}{9}\sin\frac\alpha2=\frac{SP\cdot S_1P\,\sin^2\frac\alpha2}{9}$$

The cosine rule with $S_1P-SP=2a=20$, $SS_1=2\sqrt{164}$:

$$656=400+4\,SP\cdot S_1P\,\sin^2\frac\alpha2$$$$SP\cdot S_1P\,\sin^2\frac\alpha2=64=b^2$$

Step 4. Value

$$\frac{64}{9}=7.11\ldots$$

Answer: $7$

Loci connected with the hyperbola

Method

1. Take the moving point as $(a\sec\theta,b\tan\theta)$, $\left(ct,\frac ct\right)$, or a line $y=mx+c$ with the tangency condition.
2. Write the coordinates $(h,k)$ of the required point in terms of the parameter.
3. Eliminate the parameter with $\sec^2\theta-\tan^2\theta=1$ or the condition.
4. Replace $(h,k)$ by $(x,y)$.

Standard loci

Example 56 (IITianForum). A tangent to $\frac{x^2}{9}-\frac{y^2}{4}=1$ meets the coordinate axes, and the middle point of the intercepted portion is taken. As the tangent varies, these middle points trace $\frac{A}{x^2}-\frac{4}{y^2}=B$. Find $A+B$.

▸ SHOW SOLUTION

Step 1. Tangent at θ

$$\frac x3\sec\theta-\frac y2\tan\theta=1$$

It meets the axes at $(3\cos\theta,0)$ and $(0,-2\cot\theta)$.

Step 2. Middle point

$$h=\frac32\cos\theta,\qquad k=-\cot\theta$$

Step 3. Eliminate θ

$\sec\theta=\frac{3}{2h}$, $\tan\theta=-\frac1k$:

$$\frac{9}{4h^2}-\frac{1}{k^2}=1$$$$\frac{9}{x^2}-\frac{4}{y^2}=4$$

$A=9$, $B=4$.

Answer: $13$

Example 57 (JEE Main 2022, 26 June Shift 2). Let a line $L_1$ be tangent to the hyperbola $\frac{x^2}{16}-\frac{y^2}{4}=1$ and let $L_2$ be the line passing through the origin and perpendicular to $L_1$. If the locus of the point of intersection of $L_1$ and $L_2$ is $\left(x^2+y^2\right)^2=\alpha x^2+\beta y^2$, then $\alpha+\beta$ is equal to ______.

▸ SHOW SOLUTION

Step 1. The foot (h, k)

$L_2$ is $OM$ with $M(h,k)$, so $L_1$ is perpendicular to $OM$ at $M$:

$$hx+ky=h^2+k^2$$

Step 2. Tangency

$lx+my=n$ touches when $a^2l^2-b^2m^2=n^2$:

$$16h^2-4k^2=\left(h^2+k^2\right)^2$$

$\alpha=16$, $\beta=-4$.

Answer: $12$

Example 58 (JEE Main 2021, 16 March Shift 1). The locus of the midpoints of the chord of the circle $x^2+y^2=25$ which is tangent to the hyperbola $\frac{x^2}{9}-\frac{y^2}{16}=1$ is

▸ SHOW SOLUTION

Step 1. Chord of the circle with middle point (h, k)

$T=S_1$ for the circle:

$$hx+ky=h^2+k^2$$

Step 2. Tangent to the hyperbola

$$9h^2-16k^2=\left(h^2+k^2\right)^2$$

Answer: option (4)

Example 59 (JEE Main 2021, 25 July Shift 1). The locus of the centroid of the triangle formed by any point $P$ on the hyperbola $16x^2-9y^2+32x+36y-164=0$, and its foci is

▸ SHOW SOLUTION

Step 1. Standard form

$$16(x+1)^2-9(y-2)^2=164+16-36$$$$16(x+1)^2-9(y-2)^2=144$$$$\frac{(x+1)^2}{9}-\frac{(y-2)^2}{16}=1$$

Step 2. Foci

$ae=5$: foci $(4,2)$ and $(-6,2)$, adding to $(-2,4)$.

Step 3. Centroid

$$h=\frac{x_1-2}{3},\qquad k=\frac{y_1+4}{3}$$

$x_1+1=3(h+1)$ and $y_1-2=3(k-2)$:

$$16\cdot9(h+1)^2-9\cdot9(k-2)^2=144$$$$16(h+1)^2-9(k-2)^2=16$$

Step 4. Expand

$$16x^2-9y^2+32x+36y-36=0$$

Answer: option (1)

Example 60 (IITianForum). From every point of the parabola $y^2=4x$ a pair of tangents is drawn to the ellipse $x^2+2y^2+4x+2=0$. All the chords of contact touch one fixed curve. That curve is

▸ SHOW SOLUTION

Step 1. Tangents exist

The ellipse is $(x+2)^2+2y^2=2$, with $x\le-2+\sqrt2<0$; every point of the parabola has $x\ge0$, so it is outside.

Step 2. Chord of contact from (t², 2t)

$T=0$:

$$xt^2+2y(2t)+2\left(x+t^2\right)+2=0$$$$(x+2)t^2+4yt+2(x+1)=0$$

Step 3. The fixed curve

A line of this family through a point touches the curve when the quadratic in $t$ has equal roots:

$$16y^2=8(x+2)(x+1)$$$$2y^2=x^2+3x+2$$

Step 4. Identify

$$\left(x+\frac32\right)^2-2y^2=\frac14$$$$\frac{\left(x+\frac32\right)^2}{1/4}-\frac{y^2}{1/8}=1$$$$e^2=1+\frac{1/8}{1/4}=\frac32$$

Answer: option (C)

Which method to use

What JEE asks

Across the JEE Main papers from 2017 to 2026 and the JEE Advanced and IIT JEE papers, the chapter appears in these forms.

1. Parts of the hyperbola: eccentricity, latus rectum, foci and directrices from given data, often mixed with an ellipse (Examples 3 and 5 to 15, 20, 22 and 23). This is the most common form in JEE Main: ten of the twelve 2026 questions in this note are of this type.
2. Focal distances and the focal triangle (Examples 16 to 18 and 55).
3. Line meeting the curve and tangents, including common tangents (Examples 24 to 35).
4. Normals (Examples 36 to 39).
5. Chords, chord of contact and middle points (Examples 40 to 43 and 47).
6. Asymptotes and $xy=c^2$ (Examples 2 and 44 to 51).
7. Loci (Examples 41 and 56 to 60).

JEE Advanced 2026 combined an ellipse, a confocal hyperbola and a parabola (Example 15). Earlier JEE Advanced and IIT JEE papers asked for a common tangent with a circle (Example 34), the focal directrix point on a tangent (Example 27), tangents of a given slope (Examples 29 and 30), normals (Examples 36, 38 and 39), the conjugate axis and a vertex (Example 12) and the focal triangle with the tangent (Example 55).

Quick recall

Practise these methods on real exam questions in the doMath Hyperbola chapter. The closed conic is developed in the Ellipse note, the conic with $e=1$ in the Parabola note, and tangents and normals to a general curve in the Tangents, Normals, Rolle and LMVT note.