Parabola for JEE: Notes, Focal Chords, Tangents, Normals and PYQs

A parabola is the curve of a thrown ball, the shape of a dish antenna and the graph of every quadratic. In coordinate geometry it is the simplest conic: one focus, one directrix, eccentricity exactly $1$. Almost every JEE question on it reduces to one idea, the parametric point $(at^2,2at)$, which turns chords, tangents and normals into short relations between parameters.

This note covers the focus and directrix definition, the four standard forms, shifted and slanted parabolas, the position of a point, a line meeting the parabola, chords and focal chords, chords through the vertex, chords with a given middle point, tangents in all three forms, pairs of tangents and the chord of contact, common tangents, normals and co-normal points, shortest distance, the reflection property, circles on the parabola 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 parabola

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. $S$ is the focus, $\ell$ the directrix and the constant $e$ the eccentricity.

So a parabola is the locus of a point whose distance from the focus equals its distance from the directrix:

$$PS=PM$$ A parabola opening to the right with its focus S on the axis, the directrix as a vertical line on the left, a point P on the curve joined to S and to the foot M of its perpendicular on the directrix, the two equal segments PS and PM marked

Terms used throughout.

• Axis: the line through the focus perpendicular to the directrix. The parabola is symmetric about it.
• Vertex: the point where the parabola meets its axis. It is the midpoint of the focus and the foot of the axis on the directrix.
• Focal distance of a point: its distance from the focus.
• Focal chord: a chord through the focus.
• Double ordinate: a chord perpendicular to the axis.
• Latus rectum: the focal chord perpendicular to the axis.

Recognising a parabola. The general second degree equation

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

is a parabola when $h^2=ab$ and the equation does not split into a pair of lines, that is

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

The condition $h^2=ab$ says the second degree terms form a perfect square, $(\sqrt a\,x\pm\sqrt b\,y)^2$.

Example 1 (IITianForum). Find the equation of the parabola with focus $S(0,-3)$ and directrix $y=3$. Find its vertex, axis and the length of its latus rectum.

▸ SHOW SOLUTION

Step 1. Use PS = PM

For $P(x,y)$, the distance from $y=3$ is $\lvert y-3\rvert$:

$$x^2+(y+3)^2=(y-3)^2$$$$x^2=-12y$$

Step 2. Read off the parts

The axis is the line through $S$ perpendicular to $y=3$, that is the $y$-axis $x=0$. The vertex is the midpoint of $S(0,-3)$ and the foot $(0,3)$, that is $(0,0)$. With $x^2=-4ay$, $a=3$, so the latus rectum is $4a=12$.

Answer: $x^2=-12y$, vertex $(0,0)$, axis $x=0$, latus rectum $12$

The standard parabola

Take the vertex at the origin and the axis along the $x$-axis, with focus $S(a,0)$, $a>0$, and directrix $x=-a$. For $P(x,y)$:

$$(x-a)^2+y^2=(x+a)^2$$ $$y^2=4ax$$

This is the standard parabola. Since $y^2\ge0$, every point has $x\ge0$; the curve is symmetric about the $x$-axis and opens towards the focus.

The four standard forms

The squared variable names the axis and the sign of the other term gives the direction in which the parabola opens.

Four small parabolas side by side, each with its focus marked and its directrix dashed: the first opens right, the second opens left, the third opens up and the fourth opens down

In every form the length of the latus rectum is $4a$, the eccentricity is $1$, and the focus and the directrix lie at the same distance $a$ from the vertex, on opposite sides.

Latus rectum and focal distance

Putting $x=a$ in $y^2=4ax$ gives $y=\pm2a$. So the ends of the latus rectum are

$$L(a,2a),\qquad L'(a,-2a)$$

and its length is $4a$. The half length $2a$ is the semi-latus rectum.

The parabola y squared equals 4ax with the vertex A at the origin, the focus S, the axis, the dashed directrix, and the latus rectum LL' drawn as a vertical segment through S

For a point $P(x_1,y_1)$ on $y^2=4ax$, the focal distance equals the distance from the directrix $x=-a$:

$$SP=x_1+a$$

So focal distances on $y^2=4ax$ need no square root at all.

Parametric form

For every real $t$ the point

$$x=at^2,\qquad y=2at$$

satisfies $y^2=4ax$, and every point of the parabola arises from exactly one $t$, namely $t=\frac{y}{2a}$. We call it the point $t$. In this form

$$SP=a(1+t^2)$$

Example 2 (IITianForum). For the parabola $y^2=12x$, find the focus, the directrix, the ends of the latus rectum, and the points of the parabola whose focal distance is $12$.

▸ SHOW SOLUTION

Step 1. Compare with y² = 4ax

$4a=12$, so $a=3$. Focus $(3,0)$, directrix $x=-3$, ends of the latus rectum $(3,\pm6)$.

Step 2. Focal distance 12

$SP=x+3=12$ gives $x=9$, and then

$$y^2=108,\qquad y=\pm6\sqrt3$$

Answer: focus $(3,0)$, directrix $x=-3$, $L(3,\pm6)$, points $(9,\pm6\sqrt3)$

Example 3 (JEE Main 2026, 6 April Shift 1). Let chord $PQ$ of length $3\sqrt{13}$ of the parabola $y^2=12x$ be such that the ordinates of points $P$ and $Q$ are in the ratio $1:2$. If the chord $PQ$ subtends an angle $\alpha$ at the focus of the parabola, then $\sin\alpha$ is equal to

▸ SHOW SOLUTION

Step 1. Points in parameters

$a=3$, so $P(3t^2,6t)$. The ordinate of $Q$ is twice that of $P$, so $Q$ has parameter $2t$: $Q(12t^2,12t)$.

Step 2. Use the length

$$PQ^2=(9t^2)^2+(6t)^2=117$$$$9t^4+4t^2-13=0$$$$(t^2-1)(9t^2+13)=0$$

So $t^2=1$; take $t=1$: $P(3,6)$, $Q(12,12)$. ($t=-1$ gives the mirror image and the same angle.)

Step 3. Angle at the focus

$S(3,0)$. $SP$ is vertical, of length $6$. $SQ$ joins $(3,0)$ to $(12,12)$: it rises $12$ while moving $9$ across. The angle between $SQ$ and the vertical satisfies

$$\tan\alpha=\frac{9}{12}=\frac34$$

so $\sin\alpha=\frac35$.

Answer: option (1)

General forms of the parabola

Axis parallel to a coordinate axis

Shift the vertex to $(h,k)$, keeping the axis parallel to the $x$-axis:

$$(y-k)^2=4a(x-h)$$

With the axis vertical, $(x-h)^2=4a(y-k)$. Expanding gives the familiar graph

$$y=Ax^2+Bx+C,\qquad A\ne0$$

Completing the square,

$$y-\left(-\frac{D}{4A}\right)=A\left(x+\frac{B}{2A}\right)^2$$

with $D=B^2-4AC$. So the vertex is $\left(-\frac{B}{2A},-\frac{D}{4A}\right)$, the axis is $x=-\frac{B}{2A}$, and comparing with $(x-h)^2=\frac1A(y-k)$ the latus rectum is $\frac{1}{\lvert A\rvert}$.

Example 4 (JEE Main 2026, 2 April Shift 2). Let the parabola $y=x^2+px+q$ pass through the point $(1,-1)$ such that the distance between its vertex and the $x$-axis is minimum. Then the value of $p^2+q^2$ is

▸ SHOW SOLUTION

Step 1. Use the point

$-1=1+p+q$, so $q=-2-p$.

Step 2. Ordinate of the vertex

The vertex ordinate is $\frac{4q-p^2}{4}$:

$$\frac{-8-4p-p^2}{4}=-\frac{(p+2)^2+4}{4}$$

Step 3. Least distance

The distance from the $x$-axis is $\frac{(p+2)^2+4}{4}$, least when $p=-2$. Then $q=0$ and $p^2+q^2=4$.

Answer: option (B)

Example 5 (JEE Main 2026, 5 April Shift 1). Let $P$ be a moving point on the circle $x^2+y^2-6x-8y+21=0$. Then the maximum distance of $P$ from the vertex of the parabola $x^2+6x+y+13=0$ is

▸ SHOW SOLUTION

Step 1. Vertex of the parabola

$$(x+3)^2=-(y+4)$$

So the vertex is $A(-3,-4)$; the parabola opens downwards.

Step 2. The circle

Centre $C(3,4)$, radius $\sqrt{9+16-21}=2$.

Step 3. Farthest point

The farthest point of a circle from $A$ lies on line $AC$ beyond $C$:

$$AC+r=\sqrt{36+64}+2=12$$

Answer: option (3)

Parabola from a focus and a directrix

When the directrix is slanted, write the definition directly. For focus $(\alpha,\beta)$ and directrix $lx+my+n=0$:

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

The axis is the line through the focus perpendicular to the directrix, and the vertex is the midpoint of the focus and the foot $Z$ of the axis on the directrix.

A parabola whose axis is the line y equals x: the focus S at (2,2), the vertex V at (1,1), the slanted directrix x plus y equals 0 drawn dashed, and the axis drawn through S, V and the foot Z at the origin

The two-line form. If $L_1=0$ and $L_2=0$ are perpendicular lines, then

(distance of $P$ from $L_1$)$^2=\lambda\cdot$(distance of $P$ from $L_2$)

is a parabola with axis $L_1=0$, tangent at the vertex $L_2=0$ and latus rectum $\lvert\lambda\rvert$. For $L_1:\ x-y=0$ and $L_2:\ x+y-2=0$ this reads

$$\frac{(x-y)^2}2=\lambda\,\frac{x+y-2}{\sqrt2}$$

Example 6 (JEE Main 2025, 7 April Shift 1). Let $P$ be the parabola whose focus is $(-2,1)$ and directrix is $2x+y+2=0$. Then the sum of the ordinates of the points on $P$ whose abscissa is $-2$, is

▸ SHOW SOLUTION

Step 1. The definition at x = −2

The distance from the focus is $\lvert y-1\rvert$; the distance from the directrix is $\frac{\lvert y-2\rvert}{\sqrt5}$:

$$5(y-1)^2=(y-2)^2$$

Step 2. Sum of the roots

$$4y^2-6y+1=0$$

The discriminant $36-16>0$, so both ordinates are real; their sum is $\frac64=\frac32$.

Answer: option (1)

Example 7 (JEE Main 2025, 24 January Shift 2). If the equation of the parabola with vertex $V\left(\frac32,3\right)$ and directrix $x+2y=0$ is $\alpha x^2+\beta y^2-\gamma xy-30x-60y+225=0$, then $\alpha+\beta+\gamma$ is equal to

▸ SHOW SOLUTION

Step 1. Foot of the axis on the directrix

The axis through $V$ has direction $(1,2)$. The point $\left(\frac32+u,\,3+2u\right)$ lies on $x+2y=0$ when $\frac{15}2+5u=0$, $u=-\frac32$. So $Z=(0,0)$.

Step 2. Focus

$V$ is the midpoint of $Z$ and $S$, so $S=(3,6)$.

Step 3. Equation

$$5\left[(x-3)^2+(y-6)^2\right]=(x+2y)^2$$$$4x^2+y^2-4xy-30x-60y+225=0$$

So $\alpha=4$, $\beta=1$, $\gamma=4$, and $\alpha+\beta+\gamma=9$.

Answer: option (4)

Example 8 (JEE Main 2025, 4 April Shift 2). The axis of a parabola is the line $y=x$ and its vertex and focus are in the first quadrant at distances $\sqrt2$ and $2\sqrt2$ units from the origin, respectively. If the point $(1,k)$ lies on the parabola, then a possible value of $k$ is

▸ SHOW SOLUTION

Step 1. Vertex, focus, directrix

On $y=x$ in the first quadrant: $V(1,1)$ and $S(2,2)$. The foot $Z$ is the reflection of $S$ in $V$, $Z(0,0)$, and the directrix is the line through $Z$ perpendicular to the axis: $x+y=0$.

Step 2. Equation

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

Step 3. Put x = 1

$$(1-k)^2=8(k-1)$$$$(k-1)(k-9)=0$$

So $k=1$ or $k=9$.

Answer: option (2)

Example 9 (IITianForum). Find the vertex, focus, directrix and latus rectum of the parabola

$$x^2-2xy+y^2-4x-4y+4=0$$

▸ SHOW SOLUTION

Step 1. Two-line form

The second degree part is $(x-y)^2$, so write

$$(x-y)^2=4(x+y-1)$$

The lines $x-y=0$ and $x+y-1=0$ are perpendicular.

Step 2. Distances

Divide both sides by $2$:

$$\left(\frac{x-y}{\sqrt2}\right)^2=2\sqrt2\cdot\frac{x+y-1}{\sqrt2}$$

This is (distance from $x-y=0$)$^2=2\sqrt2\,$(distance from $x+y-1=0$). So the axis is $x=y$, the tangent at the vertex is $x+y=1$ and the latus rectum is $2\sqrt2$, that is $a=\frac{1}{\sqrt2}$.

Step 3. Vertex, focus, directrix

The vertex is where $x=y$ meets $x+y=1$: $V\left(\frac12,\frac12\right)$. The parabola lies where $x+y-1\ge0$, so the focus is a distance $\frac1{\sqrt2}$ from $V$ along the axis on that side: $S(1,1)$. The directrix is parallel to $x+y=1$ at the same distance on the other side: $x+y=0$.

As a check, $2\left[(x-1)^2+(y-1)^2\right]=(x+y)^2$ expands to the given equation.

Answer: vertex $\left(\frac12,\frac12\right)$, focus $(1,1)$, directrix $x+y=0$, latus rectum $2\sqrt2$

Image of a parabola in a line

The image of a curve in a line is obtained point by point. For a line of slope $\pm1$ the reflection is simple:

So a point $(X,Y)$ lies on the image exactly when its own reflection lies on the original curve.

Example 10 (JEE Main 2026, 24 January Shift 2). Let the image of the parabola $x^2=4y$ in the line $x-y=1$ be $(y+a)^2=b(x-c)$, with $a,b,c\in\mathbb N$. Then $a+b+c$ is equal to

▸ SHOW SOLUTION

Step 1. Reflection in y = x − 1

The mirror is $y=x-1$, so the image of $(x,y)$ is $(y+1,\ x-1)$.

Step 2. Condition on the image

$(X,Y)$ is on the image when $(Y+1,\ X-1)$ is on $x^2=4y$:

$$(Y+1)^2=4(X-1)$$

So $a=1$, $b=4$, $c=1$ and $a+b+c=6$.

Answer: option (3)

Example 11 (JEE Advanced 2015, Paper 1). Let the curve $C$ be the mirror image of the parabola $y^2=4x$ with respect to the line $x+y+4=0$. If $A$ and $B$ are the points of intersection of $C$ with the line $y=-5$, then the distance between $A$ and $B$ is ______.

▸ SHOW SOLUTION

Step 1. Reflection in x + y = −4

The image of $(x,y)$ is $(-4-y,\ -4-x)$.

Step 2. Equation of C

$(X,Y)$ is on $C$ when $(-4-Y,\ -4-X)$ is on $y^2=4x$:

$$(X+4)^2=-4(Y+4)$$

Step 3. Meet y = −5

$$(X+4)^2=4,\qquad X=-2\ \text{or}\ -6$$

So $AB=4$.

Answer: $4$

Position of a point and a line meeting the parabola

Write $S\equiv y^2-4ax$ and, for a point $(x_1,y_1)$, $S_1=y_1^2-4ax_1$.

A line and the parabola. Put $y=mx+c$ ($m\ne0$) in $y^2=4ax$:

$$m^2x^2+2(mc-2a)x+c^2=0$$

Its discriminant, divided by $4$, is

$$(mc-2a)^2-m^2c^2=4a(a-mc)$$

So the condition of tangency is

$$c=\frac am$$

A line parallel to the axis ($m=0$) meets the parabola in exactly one point but is never a tangent.

Length of the chord cut by $y=mx+c$:

$$\frac{4}{m^2}\sqrt{a(1+m^2)(a-mc)}$$

Example 12 (IITianForum). For the parabola $y^2=16x$ and the line $y=2x+c$: (a) find the values of $c$ for which the line cuts, touches or misses the parabola; (b) find the length of the chord when $c=-6$; (c) state whether $(1,5)$ and $(4,7)$ lie inside or outside the parabola.

▸ SHOW SOLUTION

Step 1. Part (a)

$a=4$, $m=2$: $a-mc=4-2c$. The line cuts the parabola for $c<2$, touches it for $c=2$ and misses it for $c>2$.

Step 2. Part (b)

With $c=-6$, $a-mc=16$:

$$\frac44\sqrt{4\cdot5\cdot16}=\sqrt{320}=8\sqrt5$$

Step 3. Part (c)

$(1,5)$: $S_1=25-16>0$, outside. $(4,7)$: $S_1=49-64<0$, inside.

Answer: (a) cuts for $c<2$, touches at $c=2$, misses for $c>2$; (b) $8\sqrt5$; (c) $(1,5)$ outside, $(4,7)$ inside

Example 13 (JEE Main 2019, 10 January Shift 2). The length of the chord of the parabola $x^2=4y$ having equation $x-\sqrt2\,y+4\sqrt2=0$ is

▸ SHOW SOLUTION

Step 1. Meet the line

$y=\frac{x}{\sqrt2}+4$, so

$$x^2=2\sqrt2\,x+16$$$$x^2-2\sqrt2\,x-16=0$$

So $x=4\sqrt2$ or $x=-2\sqrt2$.

Step 2. Length

The difference of abscissae is $6\sqrt2$ and the slope is $\frac1{\sqrt2}$:

$$6\sqrt2\cdot\sqrt{1+\frac12}=6\sqrt3$$

Answer: option (1)

Chords of the parabola

Chord joining two points

The chord joining the points $t_1$ and $t_2$ of $y^2=4ax$ has slope

$$\frac{2at_1-2at_2}{at_1^2-at_2^2}=\frac{2}{t_1+t_2}$$

and equation

$$(t_1+t_2)y=2x+2at_1t_2$$

In coordinates, the chord joining $(x_1,y_1)$ and $(x_2,y_2)$ is $y(y_1+y_2)=4ax+y_1y_2$.

Focal chords

The chord through $t_1$, $t_2$ meets the axis at $(-at_1t_2,0)$. It passes through the focus $(a,0)$ exactly when

$$t_1t_2=-1$$

So if one end of a focal chord is $t$, the other end is $-\frac1t$.

The parabola y squared equals 4ax with a focal chord PQ through the focus S, P above the axis and Q below, and the angle theta that PQ makes with the axis marked at S

Why the harmonic property holds. $SQ=\frac{a(1+t^2)}{t^2}$, so

$$\frac1{SP}+\frac1{SQ}=\frac{1+t^2}{a(1+t^2)}=\frac1a$$

So the semi-latus rectum $2a$ is the harmonic mean of $SP$ and $SQ$. Also $SP\cdot SQ=a\cdot PQ$.

Why $4a\,\mathrm{cosec}^2\theta$. The slope of the focal chord is $\frac{2}{t-\frac1t}=\tan\theta$, so $t-\frac1t=2\cot\theta$ and

$$a\left(t+\frac1t\right)^2=a\left(4\cot^2\theta+4\right)$$

which is $4a\,\mathrm{cosec}^2\theta$. Since $\mathrm{cosec}^2\theta\ge1$, no focal chord is shorter than the latus rectum.

Example 14 (IITianForum). One end of a focal chord of $y^2=8x$ is $P(8,8)$. Find the other end $Q$, the length $PQ$, and check the harmonic property.

▸ SHOW SOLUTION

Step 1. Parameters

$a=2$ and $P=(2t^2,4t)$ with $t=2$. So $Q$ has parameter $-\frac12$: $Q\left(\frac12,-2\right)$.

Step 2. Length

$$PQ=a\left(t+\frac1t\right)^2=2\cdot\frac{25}{4}=\frac{25}2$$

Step 3. Focal segments

$SP=8+2=10$ and $SQ=\frac12+2=\frac52$, so

$$\frac1{10}+\frac25=\frac12=\frac1a$$

Answer: $Q\left(\frac12,-2\right)$, $PQ=\frac{25}2$, $\frac1{SP}+\frac1{SQ}=\frac12$

Example 15 (JEE Main 2025, 2 April Shift 2). Let the point $P$ of the focal chord $PQ$ of the parabola $y^2=16x$ be $(1,-4)$. If the focus of the parabola divides the chord $PQ$ in the ratio $m:n$, $\gcd(m,n)=1$, then $m^2+n^2$ is equal to

▸ SHOW SOLUTION

Step 1. Other end

$a=4$; $P(4t^2,8t)$ with $t=-\frac12$. So $Q$ has parameter $2$: $Q(16,16)$.

Step 2. Focal distances

$SP=1+4=5$ and $SQ=16+4=20$. So $S$ divides $PQ$ in the ratio $5:20=1:4$, and $m^2+n^2=17$.

Answer: option (1)

Example 16 (JEE Main 2026, 21 January Shift 2). Let one end of a focal chord of the parabola $y^2=16x$ be $(16,16)$. If $P(\alpha,\beta)$ divides this focal chord internally in the ratio $5:2$, then the minimum value of $\alpha+\beta$ is

▸ SHOW SOLUTION

Step 1. Other end

$a=4$; $A(16,16)$ has $t=2$, so the other end has parameter $-\frac12$: $B(1,-4)$.

Step 2. The two orders

The ratio $5:2$ can be taken from either end. Measured from $A$:

$$P=\frac{5B+2A}{7}=\left(\frac{37}7,\frac{12}7\right)$$

with $\alpha+\beta=7$. Measured from $B$:

$$P=\frac{5A+2B}{7}=\left(\frac{82}7,\frac{72}7\right)$$

with $\alpha+\beta=22$. The minimum is $7$.

Answer: option (2)

Example 17 (JEE Main 2026, 5 April Shift 2). Let the directrix of the parabola $P:y^2=8x$ cut the $x$-axis at the point $A$. Let $B(\alpha,\beta)$, $\alpha>1$, be a point on $P$ such that the slope of $AB$ is $\frac35$. If $BC$ is a focal chord of $P$, then six times the area of $\triangle ABC$ is

▸ SHOW SOLUTION

Step 1. Point B

$a=2$, $A(-2,0)$, $B(2t^2,4t)$:

$$\frac{4t}{2t^2+2}=\frac35$$$$3t^2-10t+3=0$$

So $t=3$ or $\frac13$. $\alpha=2t^2>1$ forces $t=3$: $B(18,12)$.

Step 2. Point C

Focal chord: $C$ has parameter $-\frac13$, $C\left(\frac29,-\frac43\right)$.

Step 3. Area

$A$ lies on the axis, so split $\triangle ABC$ by the axis. $BC$ meets the axis at the focus $S(2,0)$, and $AS=4$:

$$\text{area}=\frac12\cdot4\left(12+\frac43\right)=\frac{80}3$$

Six times the area is $160$.

Answer: option (2)

Example 18 (JEE Main 2025, 22 January Shift 2). Let $P\left(4,4\sqrt3\right)$ be a point on the parabola $y^2=4ax$ and $PQ$ be a focal chord of the parabola. If $M$ and $N$ are the feet of perpendiculars drawn from $P$ and $Q$ respectively on the directrix of the parabola, then the area of the quadrilateral $PQMN$ is equal to

▸ SHOW SOLUTION

Step 1. The parabola and Q

$48=16a$, so $a=3$. $P$ has $t=\frac{2}{\sqrt3}$, so $Q$ has $-\frac{\sqrt3}2$:

$$Q\left(\frac94,-3\sqrt3\right)$$

Step 2. A trapezium

$PM$ and $QN$ are both perpendicular to the directrix $x=-3$, so $PQNM$ is a trapezium with

$$PM=SP=7,\qquad QN=SQ=\frac{21}4$$

and height $4\sqrt3+3\sqrt3=7\sqrt3$.

Step 3. Area

$$\frac12\left(7+\frac{21}4\right)7\sqrt3=\frac{343\sqrt3}8$$

Answer: option (3)

Example 19 (JEE Advanced 2013, Paper 2). Let $PQ$ be a focal chord of the parabola $y^2=4ax$. The tangents to the parabola at $P$ and $Q$ meet at a point lying on the line $y=2x+a$, $a>0$. The length of the chord $PQ$ is

▸ SHOW SOLUTION

Step 1. Where the tangents meet

The tangents at $t_1$, $t_2$ meet at $\left(at_1t_2,\ a(t_1+t_2)\right)$. For a focal chord $t_1t_2=-1$, so the point is $\left(-a,\ a(t_1+t_2)\right)$, on the directrix.

Step 2. Use the line

On $y=2x+a$ at $x=-a$: $y=-a$. So $t_1+t_2=-1$.

Step 3. Length

$$PQ=a(t_1-t_2)^2=a\left[(t_1+t_2)^2-4t_1t_2\right]$$$$PQ=a(1+4)=5a$$

The meeting point of the tangents is proved in the section on tangents.

Answer: option (B)

Example 20 (JEE Advanced 2013, Paper 2). Let $PQ$ be a focal chord of the parabola $y^2=4ax$. The tangents to the parabola at $P$ and $Q$ meet at a point lying on the line $y=2x+a$, $a>0$. If the chord $PQ$ subtends an angle $\theta$ at the vertex of $y^2=4ax$, then $\tan\theta=$

▸ SHOW SOLUTION

Step 1. From Example 19

$t_1+t_2=-1$ and $t_1t_2=-1$, so $(t_1-t_2)^2=5$.

Step 2. Slopes from the vertex

$OP$ and $OQ$ have slopes $\frac2{t_1}$ and $\frac2{t_2}$:

$$\tan\angle POQ=\frac{\frac2{t_1}-\frac2{t_2}}{1+\frac4{t_1t_2}}=\frac{2(t_2-t_1)}{t_1t_2+4}$$

So $\lvert\tan\theta\rvert=\frac{2\sqrt5}{3}$.

Step 3. Acute or obtuse

$\vec{OP}\cdot\vec{OQ}=a^2t_1^2t_2^2+4a^2t_1t_2=-3a^2<0$, so the angle is obtuse and $\tan\theta<0$.

Answer: option (D)

Example 21 (JEE Advanced 2014, Paper 2). Let $a,r,s,t$ be nonzero real numbers. Let $P(at^2,2at)$, $Q$, $R(ar^2,2ar)$ and $S(as^2,2as)$ be distinct points on the parabola $y^2=4ax$. Suppose that $PQ$ is the focal chord and lines $QR$ and $PK$ are parallel, where $K$ is the point $(2a,0)$. The value of $r$ is

▸ SHOW SOLUTION

Step 1. Point Q

Focal chord: $Q$ has parameter $-\frac1t$.

Step 2. Equal slopes

The slope of $QR$ is $\frac{2}{r-\frac1t}$; the slope of $PK$ is $\frac{2at}{at^2-2a}=\frac{2t}{t^2-2}$. So

$$r-\frac1t=\frac{t^2-2}{t}$$$$r=\frac{t^2-1}{t}$$

Answer: option (D)

Chords that subtend a right angle at the vertex

For $P(t_1)$ and $Q(t_2)$, the slopes of $OP$ and $OQ$ are $\frac2{t_1}$ and $\frac2{t_2}$. So

$$OP\perp OQ\iff t_1t_2=-4$$

Then the chord meets the axis at $-at_1t_2=4a$: every chord that subtends a right angle at the vertex passes through the fixed point $(4a,0)$.

The parabola y squared equals 4ax with two chords OP and OQ from the vertex at right angles, and the chord PQ passing through the fixed point (4a, 0) on the axis

Example 22 (JEE Main 2026, 22 January Shift 1). If the chord joining $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ on the parabola $y^2=12x$ subtends a right angle at the vertex, then $x_1x_2-y_1y_2$ is

▸ SHOW SOLUTION

Step 1. Parameters

$a=3$: $(3t^2,6t)$, with $t_1t_2=-4$.

Step 2. Compute

$$x_1x_2-y_1y_2=9t_1^2t_2^2-36t_1t_2$$$$=9\cdot16+144=288$$

Answer: option (1)

Example 23 (JEE Main 2026, 8 April Shift 2). Let $O$ be the vertex of the parabola $y^2=4x$ and its chords $OP$ and $OQ$ be perpendicular to each other. If the locus of the mid-point of the line segment $PQ$ is a conic $C$, then the length of its latus rectum is

▸ SHOW SOLUTION

Step 1. Midpoint

$P(t_1^2,2t_1)$, $Q(t_2^2,2t_2)$ with $t_1t_2=-4$. Midpoint $(h,k)$:

$$2h=t_1^2+t_2^2,\qquad k=t_1+t_2$$

Step 2. Eliminate

$$2h=k^2-2t_1t_2=k^2+8$$

So $C$ is $y^2=2(x-4)$, with latus rectum $2$.

Answer: option (2)

Example 24 (JEE Main 2025, 23 January Shift 1). If the line $3x-2y+12=0$ intersects the parabola $4y=3x^2$ at the points $A$ and $B$, then at the vertex of the parabola, the line segment $AB$ subtends an angle equal to

▸ SHOW SOLUTION

Step 1. Points

$4\cdot\frac{3x+12}{2}=3x^2$ gives $x^2-2x-8=0$, so $A(4,12)$ and $B(-2,3)$.

Step 2. Angle at O

The slopes of $OA$ and $OB$ are $3$ and $-\frac32$:

$$\tan\theta=\left\lvert\frac{3+\frac32}{1-\frac92}\right\rvert=\frac97$$

$\vec{OA}\cdot\vec{OB}=-8+36>0$, so the angle is acute and equals $\tan^{-1}\frac97$.

Answer: option (4)

Example 25 (JEE Advanced 2015, Paper 1). Let $P$ and $Q$ be distinct points on the parabola $y^2=2x$ such that a circle with $PQ$ as diameter passes through the vertex $O$ of the parabola. If $P$ lies in the first quadrant and the area of the triangle $\triangle OPQ$ is $3\sqrt2$, then which of the following is (are) the coordinates of $P$?

▸ SHOW SOLUTION

Step 1. Right angle at O

The circle on $PQ$ passes through $O$, so $\angle POQ=90^\circ$. Write $P\left(\frac{p^2}2,p\right)$, $Q\left(\frac{q^2}2,q\right)$:

$$\frac{p^2q^2}4+pq=0,\qquad pq=-4$$

Step 2. Area

$$\triangle OPQ=\frac12\left\lvert\frac{p^2}2q-\frac{q^2}2p\right\rvert=\frac{\lvert pq\rvert\,\lvert p-q\rvert}{4}$$

So $\lvert p-q\rvert=3\sqrt2$.

Step 3. Solve

With $p>0$ and $q=-\frac4p$:

$$p+\frac4p=3\sqrt2$$$$p^2-3\sqrt2\,p+4=0$$

So $p=2\sqrt2$ or $p=\sqrt2$, giving $P\left(4,2\sqrt2\right)$ or $P\left(1,\sqrt2\right)$. Both lie in the first quadrant and both give area $3\sqrt2$.

Answer: (A) and (D)

Example 26 (IIT JEE 2010, Paper 1). Let $A$ and $B$ be two distinct points on the parabola $y^2=4x$. If the axis of the parabola touches a circle of radius $r$ having $AB$ as its diameter, then the slope of the line joining $A$ and $B$ can be

▸ SHOW SOLUTION

Step 1. Centre and radius

$A(t_1^2,2t_1)$, $B(t_2^2,2t_2)$. The centre is the midpoint, with ordinate $t_1+t_2$. The circle touches the $x$-axis, so

$$r=\lvert t_1+t_2\rvert$$

Step 2. Slope

$$m_{AB}=\frac{2}{t_1+t_2}=\pm\frac2r$$

Answer: (C) and (D)

Chord with a given middle point and diameters

Let $(h,k)$ be the middle point of a chord through $t_1$, $t_2$. Then $k=a(t_1+t_2)$, so the slope is

$$\frac{2}{t_1+t_2}=\frac{2a}{k}$$

and the chord is

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

In the $T$, $S_1$ notation, with $T\equiv yy_1-2a(x+x_1)$:

$$T=S_1$$

Diameter. The middle points of all chords of slope $m$ have $k=\frac{2a}m$. So they lie on the line $y=\frac{2a}{m}$, parallel to the axis. This line is a diameter of the parabola. The tangent at its end on the curve has slope $m$, parallel to the chords it bisects.

The parabola y squared equals 4ax with three parallel chords, their middle points marked, all lying on one horizontal line, the diameter, which meets the parabola where the tangent is parallel to the chords

Example 27 (JEE Main 2026, 21 January Shift 1). Let $O$ be the vertex of the parabola $x^2=4y$ and $Q$ be any point on it. Let $P$ divide the line segment $OQ$ internally in the ratio $2:3$, and the locus of $P$ be the conic $C$. Then the equation of the chord of $C$ which is bisected at the point $(1,2)$ is

▸ SHOW SOLUTION

Step 1. Locus

$Q(2t,t^2)$, so $P=\frac25Q=\left(\frac{4t}5,\frac{2t^2}5\right)$. With $t=\frac{5x}4$:

$$y=\frac25\cdot\frac{25x^2}{16},\qquad 5x^2=8y$$

Step 2. T = S₁ at (1, 2)

For $S\equiv5x^2-8y$: $T=5x\cdot1-4(y+2)$ and $S_1=5-16=-11$.

$$5x-4y-8=-11$$$$5x-4y+3=0$$

Answer: option (4)

Example 28 (JEE Advanced 2017, Paper 1). If a chord, which is not a tangent, of the parabola $y^2=16x$ has the equation $2x+y=p$, and midpoint $(h,k)$, then which of the following is (are) possible value(s) of $p$, $h$ and $k$?

▸ SHOW SOLUTION

Step 1. Ordinate of the midpoint

$a=4$; the slope $\frac{2a}k=\frac8k$ equals $-2$, so $k=-4$. This leaves (B) and (C).

Step 2. Midpoint on the chord

$2h+k=p$: (B) $6-4=2$ holds; (C) $4-4=0\ne-2$ fails.

Step 3. A genuine chord

The midpoint must lie inside: $k^2<16h$, that is $16<48$ for (B).

Answer: option (B)

Triangles inscribed in the parabola

With three points $t_1$, $t_2$, $t_3$ on $y^2=4ax$, the area of the triangle they form is

$$a^2\left\lvert(t_1-t_2)(t_2-t_3)(t_3-t_1)\right\rvert$$

This comes from the determinant formula after removing the common factors. The triangle formed by the tangents at the same three points has half this area.

Example 29 (JEE Main 2026, 28 January Shift 2). Let $A$ be the focus of the parabola $y^2=8x$. Let the line $y=mx+c$ intersect the parabola at two distinct points $B$ and $C$. If the centroid of triangle $ABC$ is $\left(\frac73,\frac43\right)$, then $(BC)^2$ is equal to

▸ SHOW SOLUTION

Step 1. Sum and product of parameters

$A(2,0)$, $B(2t_1^2,4t_1)$, $C(2t_2^2,4t_2)$. From the centroid:

$$4(t_1+t_2)=4,\qquad 2+2(t_1^2+t_2^2)=7$$

So $t_1+t_2=1$, $t_1^2+t_2^2=\frac52$ and $t_1t_2=-\frac34$.

Step 2. Length

$$BC^2=(t_1-t_2)^2\left[4(t_1+t_2)^2+16\right]$$

$(t_1-t_2)^2=1+3=4$, so $BC^2=4\cdot20=80$.

Answer: option (2)

Example 30 (JEE Main 2026, 23 January Shift 2). An equilateral triangle $OAB$ is inscribed in the parabola $y^2=4x$ with the vertex $O$ at the vertex of the parabola. Then the minimum distance of the circle having $AB$ as a diameter from the origin is

▸ SHOW SOLUTION

Step 1. Symmetry

The triangle is symmetric about the axis, so $A(t^2,2t)$ and $B(t^2,-2t)$, and $OA$ makes $30^\circ$ with the axis:

$$\frac{2t}{t^2}=\frac1{\sqrt3},\qquad t=2\sqrt3$$

So $A\left(12,4\sqrt3\right)$, $B\left(12,-4\sqrt3\right)$.

Step 2. The circle

Centre $(12,0)$, radius $4\sqrt3$. The origin is outside, so the least distance is

$$12-4\sqrt3=4\left(3-\sqrt3\right)$$

Answer: option (1)

Tangents

The three forms of the tangent

The tangency condition $c=\frac am$ gives the slope form:

$$y=mx+\frac am$$

touching $y^2=4ax$ at $\left(\frac a{m^2},\frac{2a}m\right)$. Every non-vertical tangent has this form, so a slope $m\ne0$ gives exactly one tangent.

Differentiating $y^2=4ax$ gives $\frac{dy}{dx}=\frac{2a}{y}$. At $(x_1,y_1)$ on the curve this gives the point form:

$$yy_1=2a(x+x_1)$$

which is $T=0$. At the point $t$ the slope is $\frac1t$, giving the parametric form:

$$ty=x+at^2$$

The tangent at the vertex is the $y$-axis, $x=0$; it is the only vertical tangent.

Example 31 (IITianForum). For the parabola $y^2=8x$ find (a) the tangent parallel to $2x-y+5=0$ and its point of contact; (b) the tangent at $(2,4)$; (c) the tangents through $(-2,-1)$.

▸ SHOW SOLUTION

Step 1. Part (a)

$a=2$, $m=2$: $y=2x+\frac22=2x+1$, touching at $\left(\frac a{m^2},\frac{2a}m\right)=\left(\frac12,2\right)$.

Step 2. Part (b)

$yy_1=2a(x+x_1)$: $4y=4(x+2)$, that is $y=x+2$.

Step 3. Part (c)

$y=mx+\frac2m$ through $(-2,-1)$:

$$-1=-2m+\frac2m$$$$2m^2-m-2=0$$$$m=\frac{1\pm\sqrt{17}}4$$

Two real slopes, as expected: $S_1=1+16>0$, so the point is outside.

Answer: (a) $y=2x+1$ at $\left(\frac12,2\right)$; (b) $y=x+2$; (c) $y=mx+\frac2m$ with $m=\frac{1\pm\sqrt{17}}4$

Where two tangents meet

Solve $t_1y=x+at_1^2$ and $t_2y=x+at_2^2$:

$$y=a(t_1+t_2),\qquad x=at_1t_2$$

So the tangents at $t_1$ and $t_2$ meet at

$$T\left(at_1t_2,\ a(t_1+t_2)\right)$$

Its abscissa is the geometric mean of the abscissae of the two points, and its ordinate is the arithmetic mean of their ordinates.

Perpendicular tangents. The slopes are $\frac1{t_1}$ and $\frac1{t_2}$, so the tangents are perpendicular when $t_1t_2=-1$. Then $T=\left(-a,\ a(t_1+t_2)\right)$ lies on the directrix, and $t_1t_2=-1$ says the chord of contact is a focal chord.

• Tangents at the ends of a focal chord are perpendicular and meet on the directrix.
• From any point of the directrix the two tangents are perpendicular.
• So the locus of the point where perpendicular tangents meet, the director circle of the parabola, is the directrix $x=-a$ itself.

The parabola y squared equals 4ax with a focal chord PQ through S, the tangents at P and Q meeting at T on the dashed directrix, and the right angle at T marked

Triangle of tangents. The tangents at $t_1$, $t_2$, $t_3$ form a triangle whose area is

$$\frac{a^2}{2}\left\lvert(t_1-t_2)(t_2-t_3)(t_3-t_1)\right\rvert$$

exactly half the area of the triangle with those three points as vertices.

Example 32 (JEE Advanced 2026, Paper 2). Let $T$ be the tangent to the parabola $y^2=16x$ at the point $(64,32)$. Let $L$ be the tangent to the same parabola at another point $(x_1,y_1)$ on the parabola. If $L$ and $T$ are perpendicular to each other, then the distance between the point $(x_1,y_1)$ and the focus of the parabola, is

▸ SHOW SOLUTION

Step 1. Parameter of (64, 32)

$a=4$; $(4t^2,8t)$ with $t=4$.

Step 2. Perpendicular tangent

Perpendicular tangents touch at the ends of a focal chord, so $(x_1,y_1)$ has parameter $-\frac14$: $\left(\frac14,-2\right)$.

Step 3. Focal distance

$$SP=x_1+a=\frac14+4=\frac{17}4$$

Answer: option (3)

Example 33 (IIT JEE 2011, Paper 1). Consider the parabola $y^2=8x$. Let $\Delta_1$ be the area of the triangle formed by the end points of its latus rectum and the point $P\left(\frac12,2\right)$ on the parabola, and $\Delta_2$ be the area of the triangle formed by drawing tangents at $P$ and at the end points of the latus rectum. Then $\frac{\Delta_1}{\Delta_2}$ is ______.

▸ SHOW SOLUTION

Step 1. Parameters

$a=2$, points $(2t^2,4t)$. $P\left(\frac12,2\right)$ has $t=\frac12$; the ends of the latus rectum $(2,\pm4)$ have $t=\pm1$.

Step 2. The two areas

$$\Delta_1=a^2\left\lvert\left(\tfrac12-1\right)(1+1)\left(-1-\tfrac12\right)\right\rvert$$$$\Delta_1=4\cdot\frac32=6$$

The tangents meet at the points $(at_it_j,\,a(t_i+t_j))$: $(1,3)$, $(-2,0)$, $(-1,-1)$. Their area is $3$, half of $\Delta_1$.

So $\frac{\Delta_1}{\Delta_2}=2$.

Answer: $2$

Standard properties of a tangent

Let the tangent at $P(at^2,2at)$ meet the axis at $T$ and let $N$ be the foot of the ordinate of $P$.

The parabola y squared equals 4ax with the tangent at P meeting the axis at T, the ordinate PN, the focus S joined to P, the foot Y of the perpendicular from S to the tangent lying on the y axis, and the equal lengths ST and SP marked

Example 34 (IITianForum). The line $x+y=8$ touches a parabola at $P(6,2)$, and $M(4,0)$ is the foot of the perpendicular drawn from $P$ to the directrix. If the vertex of the parabola is $(\alpha,\beta)$, find $\alpha+\beta$.

▸ SHOW SOLUTION

Step 1. Direction of PM

$PM$ joins $(6,2)$ and $(4,0)$, so it has direction $(1,1)$. The tangent $x+y=8$ has normal direction $(1,1)$. So $PM$ is perpendicular to the tangent: $PM$ is along the normal at $P$.

Step 2. Normal parallel to the axis

$PM$ is perpendicular to the directrix, so it is parallel to the axis. The only point where the normal is parallel to the axis is the vertex. So $P$ is the vertex.

Step 3. Check with the focus

The focus is the reflection of $M$ in $x+y=8$: $S(8,4)$. The vertex is the midpoint of $S$ and $M$: $(6,2)=P$, as found.

So $\alpha+\beta=8$.

Answer: $8$

Pair of tangents and chord of contact

From an outside point $P(x_1,y_1)$, the tangents touch at $t_1$, $t_2$ with $at_1t_2=x_1$ and $a(t_1+t_2)=y_1$. So

$$t_1+t_2=\frac{y_1}a,\qquad t_1t_2=\frac{x_1}a$$

The chord of contact joins the two points of contact. Its equation is $T=0$:

$$yy_1=2a(x+x_1)$$

The pair of tangents from $P$, as one equation, is

$$SS_1=T^2$$

with $S=y^2-4ax$, $S_1=y_1^2-4ax_1$, $T=yy_1-2a(x+x_1)$.

The triangle formed by $P$ and the two points of contact has area

$$\frac{\left(y_1^2-4ax_1\right)^{3/2}}{2a}$$

from the triangle-of-tangents formula with $(t_1-t_2)^2=\frac{y_1^2-4ax_1}{a^2}$.

Example 35 (IITianForum). A pair of tangents is drawn from $P(-2,3)$ to the parabola $y^2=8x$, touching it at $A$ and $B$. The area of the triangle $PAB$, in square units, is

▸ SHOW SOLUTION

Step 1. Parameters of A and B

$a=2$: $t_1+t_2=\frac32$, $t_1t_2=-1$. So

$$(t_1-t_2)^2=\frac94+4=\frac{25}4$$

Step 2. Area

$PAB$ is the triangle formed by the tangents at $A$, $B$ and the chord $AB$, so its area is

$$\frac{a^2}2\lvert t_1-t_2\rvert^3=2\cdot\frac{125}{8}=\frac{125}4$$

The formula $\frac{S_1^{3/2}}{2a}=\frac{125}{4}$ gives the same, with $S_1=9+16=25$.

Answer: option (A)

Example 36 (IITianForum). Tangents are drawn from a point $P$ on the directrix of $y^2=8x$, touching it at $A$ and $B$. The locus of the circumcentre of $\triangle PAB$ is

▸ SHOW SOLUTION

Step 1. A right angle at P

From a point of the directrix the tangents are perpendicular, so $\angle APB=90^\circ$ and the circumcentre is the midpoint of $AB$.

Step 2. AB is a focal chord

$A(2u^2,4u)$, $B\left(\frac2{u^2},-\frac4u\right)$. Midpoint $(h,k)$:

$$h=u^2+\frac1{u^2},\qquad k=2\left(u-\frac1u\right)$$

Step 3. Eliminate u

$$k^2=4\left(u^2+\frac1{u^2}-2\right)=4(h-2)$$

So the locus is $y^2=4x-8$.

Answer: option (A)

Common tangents

A common tangent of $y^2=4ax$ and another curve is found by writing the parabola's tangent in slope form, $y=mx+\frac am$, and imposing tangency to the other curve.

Two panels: on the left, the parabolas y squared equals 4x and x squared equals minus 32y with their common tangent touching them at (4,4) and (minus 8, minus 2); on the right, the circle x squared plus y squared equals 2 and the parabola y squared equals 8x with their two common tangents, the four points of contact forming a shaded trapezium

Example 37 (JEE Main 2014). The slope of the line touching both the parabolas $y^2=4x$ and $x^2=-32y$ is

▸ SHOW SOLUTION

Step 1. Tangent to the first parabola

$y=mx+\frac1m$.

Step 2. Tangent to x² = −32y

Put $y=mx+c$: $x^2+32mx+32c=0$. Equal roots:

$$1024m^2=128c,\qquad c=8m^2$$

Step 3. Equate

$$\frac1m=8m^2,\qquad m^3=\frac18,\qquad m=\frac12$$

Answer: option (3)

Example 38 (JEE Advanced 2014, Paper 2). The common tangents to the circle $x^2+y^2=2$ and the parabola $y^2=8x$ touch the circle at the points $P$, $Q$ and the parabola at the points $R$, $S$. Then the area of the quadrilateral $PQRS$ is

▸ SHOW SOLUTION

Step 1. Common tangents

$y=mx+\frac2m$ at distance $\sqrt2$ from the origin:

$$\frac{\lvert2/m\rvert}{\sqrt{1+m^2}}=\sqrt2$$$$m^4+m^2-2=0,\qquad m=\pm1$$

The tangents are $y=x+2$ and $y=-x-2$.

Step 2. Points of contact

On the circle (feet of the perpendiculars from $O$): $(-1,1)$ and $(-1,-1)$. On the parabola, $\left(\frac a{m^2},\frac{2a}m\right)$: $(2,4)$ and $(2,-4)$.

Step 3. Area

The quadrilateral is a trapezium with parallel sides $2$ and $8$ at distance $3$:

$$\frac{2+8}2\cdot3=15$$

Answer: option (D)

Example 39 (JEE Advanced 2018, Paper 2). Consider two straight lines, each of which is tangent to both the circle $x^2+y^2=\frac12$ and the parabola $y^2=4x$. Let these lines intersect at the point $Q$. Consider the ellipse whose center is at the origin $O(0,0)$ and whose semi-major axis is $OQ$. If the length of the minor axis of this ellipse is $\sqrt2$, then which of the following statement(s) is (are) TRUE?

▸ SHOW SOLUTION

Step 1. Common tangents

$y=mx+\frac1m$ at distance $\frac1{\sqrt2}$ from the origin:

$$\frac1{m^2(1+m^2)}=\frac12$$$$m^4+m^2-2=0,\qquad m=\pm1$$

The tangents $y=x+1$ and $y=-x-1$ meet at $Q(-1,0)$.

Step 2. The ellipse

Semi-major axis $OQ=1$, semi-minor axis $\frac1{\sqrt2}$:

$$x^2+2y^2=1$$

$e^2=1-\frac12$, so $e=\frac1{\sqrt2}$; latus rectum $\frac{2b^2}a=1$. So (A) is true and (B) is false.

Step 3. The area

$$2\int_{1/\sqrt2}^{1}\sqrt{\frac{1-x^2}2}\,dx$$

Using $\int\sqrt{1-x^2}\,dx=\frac12\left(x\sqrt{1-x^2}+\sin^{-1}x\right)$, this equals

$$\frac{\sqrt2}{2}\left(\frac\pi4-\frac12\right)=\frac{\pi-2}{4\sqrt2}$$

So (C) is true and (D) is false.

Answer: (A) and (C)

Example 40 (JEE Advanced 2015, Paper 2). Suppose that the foci of the ellipse $\frac{x^2}9+\frac{y^2}5=1$ are $(f_1,0)$ and $(f_2,0)$ where $f_1>0$ and $f_2<0$. Let $P_1$ and $P_2$ be two parabolas with a common vertex at $(0,0)$ and with foci at $(f_1,0)$ and $(2f_2,0)$, respectively. Let $T_1$ be a tangent to $P_1$ which passes through $(2f_2,0)$ and $T_2$ be a tangent to $P_2$ which passes through $(f_1,0)$. If $m_1$ is the slope of $T_1$ and $m_2$ is the slope of $T_2$, then the value of $\left(\frac1{m_1^2}+m_2^2\right)$ is ______.

▸ SHOW SOLUTION

Step 1. Foci

$c^2=9-5=4$: $f_1=2$, $f_2=-2$. So $P_1:y^2=8x$ and $P_2:y^2=-16x$.

Step 2. Tangent T₁

$y=m_1x+\frac2{m_1}$ through $(-4,0)$:

$$0=-4m_1+\frac2{m_1},\qquad m_1^2=\frac12$$

Step 3. Tangent T₂

For $y^2=-16x$ the slope form is $y=mx-\frac4m$. Through $(2,0)$:

$$0=2m_2-\frac4{m_2},\qquad m_2^2=2$$

So $\frac1{m_1^2}+m_2^2=2+2=4$.

Answer: $4$

Normals

The three forms of the normal

The normal at a point is perpendicular to the tangent there. At $(x_1,y_1)$ the tangent has slope $\frac{2a}{y_1}$, so the point form is

$$y-y_1=-\frac{y_1}{2a}(x-x_1)$$

At the point $t$ the normal has slope $-t$, giving the parametric form

$$y+tx=2at+at^3$$

Writing $m=-t$ gives the slope form

$$y=mx-2am-am^3$$

with foot $(am^2,-2am)$. So $y=mx+c$ is a normal to $y^2=4ax$ exactly when

$$c=-2am-am^3$$

Sub-normal. The normal at $P(at^2,2at)$ meets the axis at $G(2a+at^2,0)$. So $NG=2a$: the sub-normal is constant, equal to the semi-latus rectum.

Normal chord

If the normal at $P(t_1)$ meets the parabola again at $Q(t_2)$, the slope of $PQ$ is both $\frac{2}{t_1+t_2}$ and $-t_1$:

$$t_2=-t_1-\frac2{t_1}$$ The parabola y squared equals 4ax with the normal at P(t1) drawn as a chord meeting the curve again at Q(t2) on the other side of the axis

Example 41 (IITianForum). For $y^2=8x$, find the normal at $P(2,4)$, the point $Q$ where it meets the parabola again and the length $PQ$.

▸ SHOW SOLUTION

Step 1. The normal

$a=2$ and $P$ is the point $t=1$: $y+x=4+2$, that is $x+y=6$.

Step 2. The other end

$$t_2=-1-2=-3,\qquad Q(18,-12)$$

Step 3. Length

$$PQ=\sqrt{16^2+16^2}=16\sqrt2$$

Answer: $x+y=6$, $Q(18,-12)$, $PQ=16\sqrt2$

Example 42 (JEE Main 2018). Tangent and normal are drawn at $P(16,16)$ on the parabola $y^2=16x$, which intersect the axis of the parabola at $A$ and $B$, respectively. If $C$ is the centre of the circle through the points $P$, $A$ and $B$ and $\angle CPB=\theta$, then a value of $\tan\theta$ is

▸ SHOW SOLUTION

Step 1. A and B

$a=4$, $t=2$. Tangent $2y=x+16$ meets the axis at $A(-16,0)$. Normal $y+2x=16+32$ meets it at $B(24,0)$.

Step 2. The circle

$\angle APB=90^\circ$, so $AB$ is a diameter and $C=(4,0)$. Then $CP=CB$, so $\angle CPB=\angle CBP$.

Step 3. Angle at B

$BP$ goes from $(24,0)$ to $(16,16)$: up $16$, back $8$. With $BC$ along the axis,

$$\tan\theta=\frac{16}{8}=2$$

Answer: option (4)

Example 43 (JEE Advanced 2014, Paper 2). Let $a,r,s,t$ be nonzero real numbers. Let $P(at^2,2at)$, $Q$, $R(ar^2,2ar)$ and $S(as^2,2as)$ be distinct points on the parabola $y^2=4ax$. Suppose that $PQ$ is the focal chord and lines $QR$ and $PK$ are parallel, where $K$ is the point $(2a,0)$. If $st=1$, then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is

▸ SHOW SOLUTION

Step 1. The two lines

Tangent at $P$: $ty=x+at^2$. Normal at $S$, with $s=\frac1t$:

$$y+sx=2as+as^3$$

Step 2. Eliminate x

Put $x=ty-at^2$ in the normal and use $st=1$:

$$2y=\frac{2a}t+\frac a{t^3}+at$$$$y=\frac{a(t^4+2t^2+1)}{2t^3}=\frac{a(t^2+1)^2}{2t^3}$$

Answer: option (B)

Example 44 (IIT JEE 2011, Paper 2). Let $L$ be a normal to the parabola $y^2=4x$. If $L$ passes through the point $(9,6)$, then $L$ is given by

▸ SHOW SOLUTION

Step 1. Slope form through (9, 6)

$y=mx-2m-m^3$ with $a=1$:

$$6=9m-2m-m^3$$$$m^3-7m+6=0$$$$(m-1)(m-2)(m+3)=0$$

Step 2. The normals

$m=1$: $y=x-3$. $m=2$: $y=2x-12$. $m=-3$: $y=-3x+33$. These are (A), (D) and (B).

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

Example 45 (JEE Advanced 2015, Paper 1). If the normals of the parabola $y^2=4x$ drawn at the end points of its latus rectum are tangents to the circle $(x-3)^2+(y+2)^2=r^2$, then the value of $r^2$ is ______.

▸ SHOW SOLUTION

Step 1. The two normals

The ends $(1,\pm2)$ are $t=\pm1$. Normals: $y+x=3$ and $y-x=-3$.

Step 2. Distance from the centre

From $(3,-2)$:

$$\frac{\lvert3-2-3\rvert}{\sqrt2}=\sqrt2,\qquad \frac{\lvert3+2-3\rvert}{\sqrt2}=\sqrt2$$

Both equal $r$, so $r^2=2$.

Answer: $2$

Example 46 (JEE Main 2019, 10 January Shift 1). If the parabolas $y^2=4b(x-c)$ and $y^2=8ax$ have a common normal, then which one of the following is a valid choice for the ordered triad $(a,b,c)$?

▸ SHOW SOLUTION

Step 1. Normals of slope m

For $y^2=8ax$ (focal length $2a$): $y=mx-4am-2am^3$. For $y^2=4b(x-c)$, shift by $c$: $y=m(x-c)-2bm-bm^3$.

Step 2. Same line

Equal intercepts, with $m\ne0$:

$$4a+2am^2=c+2b+bm^2$$$$m^2=\frac{c+2b-4a}{2a-b}$$

Step 3. Test the triads

A real slope needs this to be positive. $(1,1,3)$ gives $\frac{1}{1}>0$; the others give $-5$, $-2$, $-2$.

The axis $y=0$ is a common normal of any two such parabolas, so the question asks for a common normal other than the axis.

Answer: option (1)

Co-normal points

Through a point $(h,k)$ the normal $y=mx-2am-am^3$ passes when

$$am^3+(2a-h)m+k=0$$

This cubic in $m$ has at most three roots, so at most three normals pass through a point. Their feet $(am_i^2,-2am_i)$ are called co-normal points.

The parabola y squared equals 4ax with three normals drawn from one point P(h, k), which lies to the right of x equals 2a, to three different feet on the curve, one above the axis and two below

Why $h>2a$. If $h\le2a$, the cubic $am^3+(2a-h)m+k$ has derivative $3am^2+(2a-h)\ge0$, so it is increasing and has only one real root.

From a point of the axis $(h,0)$, $h>2a$, the three normals are the axis itself and two with $m^2=\frac{h-2a}{a}$.

Example 47 (IITianForum). Three normals are drawn from $(11,6)$ to the parabola $y^2=4x$, with feet $\left(m_i^2,-2m_i\right)$. Find $\sum\frac1{m_i}$ and the centroid of the triangle formed by the feet.

▸ SHOW SOLUTION

Step 1. The cubic

$a=1$, $h=11$, $k=6$:

$$m^3-9m+6=0$$

Here $h>2a$; the cubic changes sign three times ($m=-4,0,1,3$ give $-22$, $6$, $-2$, $6$), so all three normals are real.

Step 2. Sum of reciprocals

$\sum m_i=0$, $\sum m_im_j=-9$, $m_1m_2m_3=-6$:

$$\sum\frac1{m_i}=\frac{\sum m_im_j}{m_1m_2m_3}=\frac{-9}{-6}=\frac32$$

Step 3. Centroid

$$\sum m_i^2=0-2(-9)=18$$

The centroid is $\left(\frac{18}3,0\right)=(6,0)$.

Answer: $\sum\frac1{m_i}=\frac32$, centroid $(6,0)$

Example 48 (IITianForum). Let $(h,k)$ be a point on the axis of the parabola $y^2-4y-4x+8=0$ from which three distinct real normals can be drawn to the parabola. Find the least possible value of $[h]$, where $[\cdot]$ is the greatest integer function.

▸ SHOW SOLUTION

Step 1. Standard form

$$(y-2)^2=4(x-1)$$

Vertex $(1,2)$, $a=1$, axis $y=2$, so $k=2$.

Step 2. Shift to the vertex

With $X=x-1$, $Y=y-2$ the point is $(h-1,0)$. Three real normals from a point of the axis need

$$h-1>2a=2,\qquad h>3$$

Step 3. Least value of [h]

$h$ can be any number above $3$, for example $3.1$, so the least value of $[h]$ is $3$.

Answer: $3$

Example 49 (IITianForum). From a point $P(h,k)$ on the parabola $y^2=8x$, two normals are drawn to the parabola at two distinct points $Q$ and $R$, both different from $P$. Find the least integral value of $h$.

▸ SHOW SOLUTION

Step 1. Normals through a point of the curve

$a=2$, $P(2s^2,4s)$. The normal at $t$ passes through $P$ when

$$4s+2s^2t=4t+2t^3$$$$(t-s)(t^2+st+2)=0$$

Step 2. Two more feet

$t=s$ is the normal at $P$ itself. Two more real, distinct feet need

$$s^2-8>0$$

and neither root equals $s$, which holds since $2s^2+2\ne0$.

Step 3. Least integral h

$h=2s^2>16$, so the least integer is $17$.

Answer: $17$

Shortest distance

The shortest distance from a point, or from another curve, to the parabola is measured along a line that is normal to the parabola. For two curves, it lies along a common normal.

Example 50 (JEE Main 2025, 23 January Shift 2). Let the shortest distance from $(a,0)$, $a>0$, to the parabola $y^2=4x$ be $4$. Then the equation of the circle passing through the point $(a,0)$ and the focus of the parabola, and having its centre on the axis of the parabola, is

▸ SHOW SOLUTION

Step 1. Which normal

If $a\le2$ the nearest point is the vertex and the distance is $a\le2$, not $4$. So $a>2$ and the nearest points are the feet $(m^2,-2m)$ with $m^2=a-2$.

Step 2. Use the distance

$$(a-m^2)^2+4m^2=4+4m^2=16$$

So $m^2=3$ and $a=5$.

Step 3. The circle

It passes through $(5,0)$ and $(1,0)$ with centre on the $x$-axis: centre $(3,0)$, radius $2$:

$$x^2+y^2-6x+5=0$$

Answer: option (1)

Example 51 (JEE Advanced 2016, Paper 2). Let $P$ be the point on the parabola $y^2=4x$ which is at the shortest distance from the center $S$ of the circle $x^2+y^2-4x-16y+64=0$. Let $Q$ be the point on the circle dividing the line segment $SP$ internally. Then

▸ SHOW SOLUTION

Step 1. The circle

Centre $S(2,8)$, radius $\sqrt{4+64-64}=2$.

Step 2. Normal through S

$y=mx-2m-m^3$ through $(2,8)$:

$$8=2m-2m-m^3,\qquad m=-2$$

So $P=(m^2,-2m)=(4,4)$ and the normal is $y=-2x+12$.

Step 3. Check each option

$SP=\sqrt{4+16}=2\sqrt5$, so (A) is true. $SQ=2$ and $QP=2\sqrt5-2$, so $SQ:QP=1:\left(\sqrt5-1\right)=\left(\sqrt5+1\right):4$, and (B) is false. The normal meets $y=0$ at $x=6$, so (C) is true. The radius $SQ$ lies along $SP$, of slope $-2$, so the tangent at $Q$ has slope $\frac12$ and (D) is true.

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

Reflection property and circles on the parabola

Reflection property

The tangent at $P$ bisects the angle between $PS$ and $PM$, where $PM$ is parallel to the axis. So the normal at $P$ makes equal angles with $PS$ and with the line through $P$ parallel to the axis. By the law of reflection:

• A ray travelling parallel to the axis is reflected through the focus.
• A ray from the focus is reflected parallel to the axis.

This is why dish antennas, torch reflectors and solar cookers have parabolic cross-sections.

The parabola y squared equals 4ax with four horizontal rays coming in from the right, each reflected at the curve so that all four reflected rays pass through the focus S

Example 52 (IITianForum). A ray along the line $y=4$, travelling towards the left, strikes the parabola $y^2=8x$ at $A$. After reflection it meets the axis of the parabola at $B$, and the normal at $A$ meets the axis at $C$. Which of the following are correct?

▸ SHOW SOLUTION

Step 1. Point A and point B

$y=4$ gives $x=2$, so $A(2,4)$. The ray is parallel to the axis, so after reflection it passes through the focus: $B=(2,0)$.

Step 2. Point C

$a=2$ and $A$ is the point $t=1$. The normal $x+y=6$ meets the axis at $C(6,0)$. (Also, the sub-normal is $2a=4$, so $C$ is $4$ to the right of the foot $(2,0)$.)

Step 3. Triangle

$AB=4$ is vertical and $BC=4$ lies on the axis, so the area is $\frac12\cdot4\cdot4=8$. $AC=4\sqrt2\ne AB$.

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

Circles on the parabola

Two panels: on the left, a focal chord PQ of the parabola with the circle on PQ as diameter touching the dashed directrix; on the right, a focal radius SP with the circle on SP as diameter touching the y axis

Example 53 (JEE Main 2025, 2 April Shift 1). Let the focal chord $PQ$ of the parabola $y^2=4x$ make an angle of $60^\circ$ with the positive $x$-axis, where $P$ lies in the first quadrant. If the circle, whose one diameter is $PS$, $S$ being the focus of the parabola, touches the $y$-axis at the point $(0,\alpha)$, then $5\alpha^2$ is equal to

▸ SHOW SOLUTION

Step 1. Point P

$S(1,0)$. Slope $\frac{2}{t-\frac1t}=\sqrt3$ with $t>0$:

$$\sqrt3t^2-2t-\sqrt3=0,\qquad t=\sqrt3$$

So $P\left(3,2\sqrt3\right)$.

Step 2. Circle on SP

Its centre is $\left(2,\sqrt3\right)$ and it touches the tangent at the vertex, the $y$-axis, at the level of the centre: $\alpha=\sqrt3$. So $5\alpha^2=15$.

Answer: option (1)

Example 54 (JEE Main 2025, 29 January Shift 2). Let $y^2=12x$ be the parabola and $S$ be its focus. Let $PQ$ be a focal chord of the parabola such that $(SP)(SQ)=\frac{147}4$. Let $C$ be the circle described taking $PQ$ as a diameter. If the equation of a circle $C$ is $64x^2+64y^2-\alpha x-64\sqrt3\,y=\beta$, then $\beta-\alpha$ is equal to ______.

▸ SHOW SOLUTION

Step 1. The parameter

$a=3$; $SP=3(1+t^2)$ and $SQ=3\left(1+\frac1{t^2}\right)$:

$$SP\cdot SQ=9\left(t+\frac1t\right)^2=\frac{147}4$$

So $\left(t+\frac1t\right)^2=\frac{49}{12}$ and $\left(t-\frac1t\right)^2=\frac{49}{12}-4=\frac1{12}$.

Step 2. Circle on PQ

$x_1+x_2=3\left(t^2+\frac1{t^2}\right)=3\left(\frac{49}{12}-2\right)=\frac{25}4$, $x_1x_2=a^2=9$, $y_1y_2=-4a^2=-36$, and $y_1+y_2=6\left(t-\frac1t\right)=\pm\sqrt3$. The circle is

$$x^2+y^2-\frac{25}4x-(y_1+y_2)y-27=0$$

Step 3. Match the given form

Multiplying by $64$, the $y$ term is $-64\sqrt3\,y$ when $y_1+y_2=\sqrt3$:

$$64x^2+64y^2-400x-64\sqrt3\,y=1728$$

So $\alpha=400$, $\beta=1728$ and $\beta-\alpha=1328$.

Answer: $1328$

Example 55 (IIT JEE 2012, Paper 1). Let $S$ be the focus of the parabola $y^2=8x$ and let $PQ$ be the common chord of the circle $x^2+y^2-2x-4y=0$ and the given parabola. The area of the triangle $PQS$ is ______.

▸ SHOW SOLUTION

Step 1. Points of intersection

Put $x=\frac{y^2}8$ in the circle:

$$\frac{y^4}{64}+y^2-\frac{y^2}4-4y=0$$$$y\left(y^3+48y-256\right)=0$$

$y^3+48y-256$ is increasing and vanishes at $y=4$, so the real roots are $y=0$ and $y=4$: $P(0,0)$, $Q(2,4)$.

Step 2. Area

$S(2,0)$; $QS$ is vertical of length $4$ and $P$ is $2$ away from it:

$$\frac12\cdot4\cdot2=4$$

Answer: $4$

Which method to use

A decision chart: from the question asked, arrows lead to eight situations, each paired with its method: focus and directrix given, distance equal to distance; general form, complete the square; two points on the curve, parameters t1 and t2; focal chord, t1 t2 equals minus 1; tangent condition, c equals a over m; point outside, pair of tangents and chord of contact; normal through a point, cubic in m; 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. Focal chords: length, harmonic property, quadrilaterals on the directrix, points dividing the chord (Examples 14 to 21). This is the most common form in JEE Main.
2. General and slanted parabolas: vertex, focus and latus rectum from a focus and directrix, or from a mirror image (Examples 4 to 11).
3. Chords with a condition: right angle at the vertex, chords with a given mid-point, inscribed triangles (Examples 22 to 30).
4. Tangents and common tangents, often with a circle or an ellipse (Examples 31 to 40). This is the common form in JEE Advanced.
5. Normals: normal chord, co-normal points, shortest distance (Examples 41 to 51).
6. Loci of mid-points and centroids (Examples 56 to 60).

JEE Advanced 2026 asked for the focal distance of the point whose tangent is perpendicular to a given tangent (Example 32). Earlier JEE Advanced and IIT JEE papers used the mirror image of a parabola (Example 11), focal chord tangents (Examples 19 and 20), the circle on a chord (Example 25), common tangents with a circle (Examples 38 and 39), normals through a point (Example 44) and a centroid locus (Example 59).

Quick recall

Practise these methods on real exam questions in the doMath Parabola chapter. Circles touching a parabola and the family of circles are developed in the Family of Circles note, tangents to a general curve in the Tangents, Normals, Rolle and LMVT note, and the discriminant conditions used for a line meeting the curve in the Quadratic Equations note.