JEE Mains Maths · Conic Sections

Parabola and Its Focal Chords

A parabola is the set of points equally far from a focus and a directrix. This page reads its vertex, focus and latus rectum from any equation, uses the parametric point to handle chords, and uses the focal-chord rules t₁t₂ = −1 and SP = a + x.

All Conic Sections notes JEE Mains Maths strategy

Why this matters

Forty-three PYQs. Most are solved by writing points as (at², 2at) and using one relation between the parameters: −1 for a focal chord, −4 for a right angle at the vertex. Six ideas cover the page.

Concept 1 of 6: Standard and shifted forms: vertex, focus, directrix, latus rectum

y^2=4ax

the number

a

is the distance from the vertex to the focus, and everything else follows from it: the directrix is

a

behind the vertex and the latus rectum, the chord through the focus perpendicular to the axis, has length

4a

. Any parabola with a horizontal or vertical axis becomes this shape after completing the square, with the vertex moved to

(h,k)

Definition

$y^2=4ax$ : vertex $(0,0)$ , focus $(a,0)$ , directrix $x=-a$ , latus rectum $4a$ with ends $(a,\pm2a)$ .
$x^2=4ay$ : focus $(0,a)$ , directrix $y=-a$ .
Shifted: $(y-k)^2=4a(x-h)$ has vertex $(h,k)$ , focus $(h+a,k)$ , directrix $x=h-a$ .
$y=px^2+qx+r$ : complete the square; $4a=\frac{1}{|p|}$ .
A point's distance from the focus equals its distance from the directrix.

The standard parabola

y^2=4ax:\quad S=(a,0),\ \ x=-a,\ \ \text{LR}=4a

Worked example

Find the vertex, focus, directrix and latus rectum of

y^2-4y-8x+20=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · 28 June 2022 · Q165Moderate

Example 1 · Conic Sections · Parabola and Its Focal Chords

If vertex of a parabola is

(2, - 1)

and the equation of its directrix is

4x - 3y = 21

, then the length of its latus rectum is

Shift the focus along with the vertex

For

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

the focus is

(h+a,k)

, not

(a,0)

. Every feature moves with the vertex.

Drill 8 more on standard and shifted forms: vertex, focus, directrix, latus rectum

Concept 2 of 6: Parabolas in any position: the focus-directrix definition

When the axis is slanted, no standard form fits. Go back to the definition: a point

P

is on the parabola exactly when its distance to the focus equals its distance to the directrix. Squaring that gives the equation directly,

xy

term and all. The geometry also gives shortcuts: the axis passes through the focus perpendicular to the directrix, and the vertex is halfway between the focus and the directrix.

Definition

Equation: $(x-\alpha)^2+(y-\beta)^2=\frac{(lx+my+n)^2}{l^2+m^2}$ for focus $(\alpha,\beta)$ , directrix $lx+my+n=0$ .
The axis is the perpendicular from the focus to the directrix.
The vertex is the midpoint of the focus and the foot of that perpendicular.
Latus rectum $=2\times$ (focus to directrix) $=4\times$ (vertex to directrix).

Focus-directrix equation

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

Worked example

Find the parabola with focus

(1,1)

and directrix

x+y=0

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q132Moderate

Example 2 · Conic Sections · Parabola and Its Focal Chords

The axis of a parabola is the line

y=x

and its vertex and focus are in the first quadrant at distances

\sqrt{2}

and

2\sqrt{2}

units from the origin, respectively. If the point

(1,k)

lies on the parabola, then a possible value of

k

is :-

Keep the $\sqrt{l^2+m^2}$

The distance to the directrix is

\frac{|lx+my+n|}{\sqrt{l^2+m^2}}

. Dropping the denominator gives a different curve.

Drill 4 more on parabolas in any position: the focus-directrix definition

Concept 3 of 6: The parametric point and chords seen from the vertex

Every point of

y^2=4ax

(at^2,2at)

for one number

t

. A chord is then fixed by two numbers

t_1,t_2

, and its slope is simply

\frac{2}{t_1+t_2}

. A chord whose ends make a right angle at the vertex satisfies

t_1t_2=-4

, and all such chords pass through

(4a,0)

Definition

Point: $(at^2,\,2at)$ .
Chord through $t_1,t_2$ : slope $\frac{2}{t_1+t_2}$ ; equation $2x-(t_1+t_2)y+2at_1t_2=0$ .
Right angle at the vertex: $t_1t_2=-4$ ; the chord passes through $(4a,0)$ .
Equilateral triangle with one vertex at the vertex: the other two are symmetric about the axis; side $8\sqrt3\,a$ .

Chord joining t₁ and t₂

2x-(t_1+t_2)y+2at_1t_2=0

Worked example

y^2=8x

, find the slope of the chord joining the points with

t=1

and

t=-2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 1 · Q67Moderate

Example 3 · Conic Sections · Parabola and Its Focal Chords

If the chord joining the points

P_{1}\left( x_{1},y_{1} \right)

and

P_{2}\left( x_{2} \right.

y_{2}

) on the parabola

y^{2}= 12x

subtends a right angle at the vertex of the parabola, then

x_{1}x_{2}-y_{1}y_{2}

is equal to

$-4$ is for the vertex, $-1$ is for the focus

A right angle at the vertex gives

t_1t_2=-4

. A chord through the focus gives

t_1t_2=-1

. Swapping them is the most common slip on this page.

Drill 6 more on the parametric point and chords seen from the vertex

Concept 4 of 6: Focal chords and focal distances

A chord through the focus has ends with

t_1t_2=-1

. The distance from the focus to a point is its distance to the directrix,

a+x

, so focal lengths need no square roots. Put together, a focal chord's length is

a\left(t+\frac1t\right)^2

, or

\frac{4a}{\sin^2\theta}

in terms of its angle with the axis.

Definition

Focal chord: $t_1t_2=-1$ ; the other end of $t$ is $-\frac1t$ .
Focal distance: $SP=a+x_P=a(1+t^2)$ .
Length: $a\left(t+\frac1t\right)^2=x_1+x_2+2a=\frac{4a}{\sin^2\theta}$ .
Harmonic property: $\frac{1}{SP}+\frac{1}{SQ}=\frac1a$ . Also $SP\cdot SQ=a\cdot PQ$ .
The latus rectum is the shortest focal chord.

Focal chord at angle θ to the axis

t_1t_2=-1,\qquad PQ=\frac{4a}{\sin^2\theta}

Worked example

One end of a focal chord of

y^2=12x

(12,12)

. Find the other end and the chord's length.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 2 Apr 2025 · Q142Moderate

Example 4 · Conic Sections · Parabola and Its Focal Chords

Let the point

P

of the focal chord

PQ

of the parabola

y^{2}= 16x

(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 :

$\theta$ is the angle with the AXIS

\frac{4a}{\sin^2\theta}

uses the chord's angle with the parabola's axis. For

x^2=4ay

the axis is vertical, so measure from the

y

-axis.

Drill 11 more on focal chords and focal distances

Concept 5 of 6: Chords of a parabola by their midpoint, and where a line meets it

For

y^2=4ax

, subtracting the equations at the two ends gives

(y_1-y_2)(y_1+y_2)=4a(x_1-x_2)

, so a chord's slope is

\frac{4a}{y_1+y_2}=\frac{2a}{k}

where

k

is the midpoint's

y

. The whole chord is

T=S_1

. When a line meets the parabola, substitute and use the sum and product of the roots, not the roots themselves.

Definition

Slope of the chord with midpoint $(h,k)$ : $\frac{2a}{k}$ .
Chord with midpoint $(h,k)$ : $T=S_1$ , i.e. $ky-2a(x+h)=k^2-4ah$ .
A line meets the parabola: substitute, then use Vieta for the sum and product of the $y$ 's (or $x$ 's).
Chord length along a line of slope $m$ : $\sqrt{1+m^2}\,|x_1-x_2|$ .

Slope of a chord of y² = 4ax with midpoint (h, k)

m=\frac{2a}{k}

Worked example

Find the chord of

y^2=12x

whose midpoint is

(3,2)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 29 January 2024 · Q173Moderate

Example 5 · Conic Sections · Parabola and Its Focal Chords

Let

P(\alpha,\beta)

be a point on the parabola

y^{2}= 4x

. If

P

also lies on the chord of the parabola

x^{2}= 8y

whose mid point is

\left( 1,\frac{5}{4} \right)

. Then

(\alpha- 28)(\beta- 8)

is equal to

Use the MIDPOINT's ordinate

The slope

\frac{2a}{k}

uses the

y

-coordinate of the midpoint. Plugging in an end's

y

gives the tangent's slope at that end instead.

Drill 4 more on chords of a parabola by their midpoint, and where a line meets it

Concept 6 of 6: Loci from a moving point on a parabola

Write the moving point as

(at^2,2at)

, express the new point in terms of

t

, then remove

t

. Because

x

depends on

t^2

and

y

t

, midpoints and centroids built this way usually give another parabola, with its own vertex and latus rectum to read off.

Definition

Put $P=(at^2,2at)$ .
Write the locus point $(h,k)$ in terms of $t$ .
Solve for $t$ from the simpler coordinate (usually $k$ ) and substitute in the other.
Rename $h,k$ as $x,y$ ; read the vertex and latus rectum of the new curve if asked.

Parametric point to eliminate

P=(at^2,\,2at)

Worked example

Find the locus of the midpoint of the segment joining the vertex of

y^2=8x

to a point of the parabola.

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 1 · Q61Moderate

Example 6 · Conic Sections · Parabola and Its Focal Chords

The locus of the mid-point of the line segment joining the focus of the parabola

y^{2}= 4ax

a

moving point of the parabola, is another parabola whose directrix is: .

Read the NEW curve's features

After finding the locus, questions ask for its latus rectum or directrix. Rewrite it in standard form first; the original parabola's

a

no longer applies.

Drill 4 more on loci from a moving point on a parabola

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (6)

Standard and shifted forms: vertex, focus, directrix, latus rectum
The standard parabola
$y^2=4ax:\quad S=(a,0),\ \ x=-a,\ \ \text{LR}=4a$
Parabolas in any position: the focus-directrix definition
Focus-directrix equation
$(x-\alpha)^2+(y-\beta)^2=\frac{(lx+my+n)^2}{l^2+m^2}$
The parametric point and chords seen from the vertex
Chord joining t₁ and t₂
$2x-(t_1+t_2)y+2at_1t_2=0$
Focal chords and focal distances
Focal chord at angle θ to the axis
$t_1t_2=-1,\qquad PQ=\frac{4a}{\sin^2\theta}$
Chords of a parabola by their midpoint, and where a line meets it
Slope of a chord of y² = 4ax with midpoint (h, k)
$m=\frac{2a}{k}$
Loci from a moving point on a parabola
Parametric point to eliminate
$P=(at^2,\,2at)$

Watch out for (6)

Shift the focus along with the vertex
→ Standard and shifted forms: vertex, focus, directrix, latus rectum
Keep the $\sqrt{l^2+m^2}$
→ Parabolas in any position: the focus-directrix definition
$-4$ is for the vertex, $-1$ is for the focus
→ The parametric point and chords seen from the vertex
$\theta$ is the angle with the AXIS
→ Focal chords and focal distances
Use the MIDPOINT's ordinate
→ Chords of a parabola by their midpoint, and where a line meets it
Read the NEW curve's features
→ Loci from a moving point on a parabola

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