JEE Mains Maths · Conic Sections

Ellipse: Axes, Eccentricity and Focal Distances

The ellipse through its numbers a, b and e: axis lengths, foci, directrices and latus rectum; ellipses with a vertical axis or a moved centre; the constant sum of focal distances; and the parametric point with the auxiliary circle.

All Conic Sections notes JEE Mains Maths strategy

Why this matters

Forty-three PYQs. Most give two facts, such as an eccentricity and a latus rectum, and ask for a third. The relation b² = a²(1 − e²) links them all. Four ideas cover the page.

Concept 1 of 4: a, b, e and the latus rectum

An ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

with

a>b

is fixed by two numbers. The eccentricity

e

measures how stretched it is: the foci sit at

(\pm ae,0)

, and

b^2=a^2(1-e^2)

ties the three together. Almost every question gives two of

a

b

e

, the focal distance

2ae

, or the latus rectum

\frac{2b^2}{a}

, and asks for another.

Definition

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

Major axis $2a$ along $x$ , minor axis $2b$ .
$b^2=a^2(1-e^2)$ , with $0<e<1$ .
Foci $(\pm ae,0)$ , distance between them $2ae$ .
Directrices $x=\pm\frac{a}{e}$ .
Latus rectum $\frac{2b^2}{a}$ .

The linking relation and the latus rectum

b^2=a^2(1-e^2),\qquad \text{LR}=\frac{2b^2}{a}

Worked example

An ellipse

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1

(

a>b

) has

e=\frac35

and latus rectum

\frac{32}{5}

. Find

a

and

b

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 24 Jan 2026 Shift 2 · Q52Moderate

Example 1 · Conic Sections · Ellipse: Axes, Eccentricity and Focal Distances

Let the length of the latus rectum of an ellipse

\frac{x^{2}}{a^{2}}+\frac{y^{2}}{{\text{ }b}^{2}}= 1,(a > b)

, be 30 . If its eccentricity is the maximum value of the function

f(t) = -\frac{3}{4}+ 2t -t^{2}

, then

\left( a^{2}+b^{2} \right)

is equal to -

$1-e^2$ for an ellipse, $e^2-1$ for a hyperbola

For an ellipse

b^2=a^2(1-e^2)

and

e<1

. Writing

a^2(e^2-1)

, the hyperbola rule, makes

b^2

negative.

Drill 13 more on a, b, e and the latus rectum

Concept 2 of 4: A vertical major axis, or a moved centre

Nothing forces the major axis onto the

x

-axis. If the bigger denominator is under

y^2

, the ellipse is tall: the foci, directrices and latus rectum all turn by

90^\circ

, and the roles of

a

and

b

swap in every formula. If the centre is at

(h,k)

, complete the squares and measure every feature from that centre.

Definition

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ with $b>a$ : major axis on $y$ , $a^2=b^2(1-e^2)$ , foci $(0,\pm be)$ , latus rectum $\frac{2a^2}{b}$ .
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ : centre $(h,k)$ ; foci, vertices and directrices shift with it.
Given the centre, a focus and a vertex on one line: $ae$ = centre to focus, semi-major = centre to vertex.

Tall ellipse (b > a)

a^2=b^2(1-e^2),\quad S=(0,\pm be),\quad \text{LR}=\frac{2a^2}{b}

Worked example

Find

e

, the foci and the latus rectum of

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

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2025 · 4 April 2025 · Q67Moderate

Example 2 · Conic Sections · Ellipse: Axes, Eccentricity and Focal Distances

The length of the latus-rectum of the ellipse, whose foci are

(2,5)

and

(2, - 3)

and eccentricity is

\frac{4}{5}

, is.

Check which denominator is bigger

For

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

, using

b^2=a^2(1-e^2)

with

a^2=9

gives

1-e^2>1

, which is impossible. The larger denominator always plays the semi-major role.

Drill 7 more on a vertical major axis, or a moved centre

Concept 3 of 4: Focal distances: the constant sum

For any point

P

on the ellipse,

SP+S'P=2a

. That is the ellipse's defining property, and it works backwards too: points whose distances from two fixed points add to a constant form an ellipse. Each focal distance is also

e

times the distance to the matching directrix, which gives

SP=a-ex

and

S'P=a+ex

without square roots.

Definition

$SP=a-ex_1$ , $S'P=a+ex_1$ for $P(x_1,y_1)$ , $S=(ae,0)$ .
$SP+S'P=2a$ .
$SP\cdot S'P=a^2-e^2x_1^2$ , between $b^2$ and $a^2$ .
Directrix property: $SP=e\cdot PM$ , $PM$ the distance to the directrix $x=\frac{a}{e}$ .
If $\angle SPS'=90^\circ$ : $SP^2+S'P^2=(2ae)^2$ .

Focal distances

SP=a-ex_1,\quad S'P=a+ex_1,\quad SP+S'P=2a

Worked example

\frac{x^2}{25}+\frac{y^2}{16}=1

, find the focal distances of the point with

x=3

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q66Moderate

Example 3 · Conic Sections · Ellipse: Axes, Eccentricity and Focal Distances

Let

S

and

S^{'}

be the foci of the ellipse

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

and

P(\alpha,\beta)

be a point on the ellipse in the first quadrant. If

(SP)^{2}+\left( S^{'}P \right)^{2}- SP \bullet S^{'}P = 37

, then

\alpha^{2}+\beta^{2}

is equal to :

$a-ex$ is the distance to the focus on the SAME side

For

S=(ae,0)

SP=a-ex_1

. The focus at

(-ae,0)

gives

a+ex_1

. Swapping them matters when only one focal distance is asked.

Drill 8 more on focal distances: the constant sum

Concept 4 of 4: The parametric point, the auxiliary circle and loci

Every point of the ellipse is

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

. Directly above it, on the circle

x^2+y^2=a^2

(the auxiliary circle), is

(a\cos\theta,\ a\sin\theta)

: the ellipse is that circle squashed vertically by

\frac{b}{a}

. That is why its area is

\pi ab

, and why midpoints and dividing points built from it trace smaller ellipses.

Definition

Parametric point: $(a\cos\theta,\ b\sin\theta)$ , $\theta$ the eccentric angle.
Auxiliary circle: $x^2+y^2=a^2$ ; the ellipse point is the circle point scaled by $\frac{b}{a}$ vertically.
Area: $\pi ab$ .
Loci: write the new point in $\theta$ , isolate $\cos\theta$ and $\sin\theta$ , square and add.
$p\cos\theta+q\sin\theta$ is at most $\sqrt{p^2+q^2}$ .

Parametric point and area

(a\cos\theta,\ b\sin\theta),\qquad \text{Area}=\pi ab

Worked example

Find the locus of the midpoint of the segment joining

(2,0)

to a point of

\frac{x^2}{16}+\frac{y^2}{9}=1

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 1 February 2024 · Q155Moderate

Example 4 · Conic Sections · Ellipse: Axes, Eccentricity and Focal Distances

Let

P

be a point on the ellipse

\frac{x^{2}}{9}+\frac{y^{2}}{4}= 1

. Let the line passing through

P

and parallel to

y

-axis meet the circle

x^{2}+y^{2}= 9

at point

Q

such that

P

and

Q

are on the same side of the

x

-axis. Then, the eccentricity of the locus of the point

R

PQ

such that

PR:RQ = 4:3

P

moves on the ellipse, is :

The auxiliary circle has radius $a$ , the SEMI-MAJOR axis

For a tall ellipse (

b>a

) the auxiliary circle is

x^2+y^2=b^2

. Always use the larger semi-axis.

Drill 11 more on the parametric point, the auxiliary circle and loci

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 (4)

a, b, e and the latus rectum
The linking relation and the latus rectum
$b^2=a^2(1-e^2),\qquad \text{LR}=\frac{2b^2}{a}$
A vertical major axis, or a moved centre
Tall ellipse (b > a)
$a^2=b^2(1-e^2),\quad S=(0,\pm be),\quad \text{LR}=\frac{2a^2}{b}$
Focal distances: the constant sum
Focal distances
$SP=a-ex_1,\quad S'P=a+ex_1,\quad SP+S'P=2a$
The parametric point, the auxiliary circle and loci
Parametric point and area
$(a\cos\theta,\ b\sin\theta),\qquad \text{Area}=\pi ab$

Watch out for (4)

$1-e^2$ for an ellipse, $e^2-1$ for a hyperbola
→ a, b, e and the latus rectum
Check which denominator is bigger
→ A vertical major axis, or a moved centre
$a-ex$ is the distance to the focus on the SAME side
→ Focal distances: the constant sum
The auxiliary circle has radius $a$ , the SEMI-MAJOR axis
→ The parametric point, the auxiliary circle and loci

Test yourself on Conic Sections

20 past JEE Mains questions from this chapter, timed at 48 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

Take the Conic Sections chapter test Drill all 43 questions