JEE Mains Maths · Conic Sections

Hyperbola: Axes, Eccentricity and Focal Distances

The hyperbola through its numbers a, b and e: axes, foci, directrices, latus rectum and the conjugate hyperbola; problems that pair a hyperbola with an ellipse; focal distances; and the rectangular hyperbola.

All Conic Sections notes JEE Mains Maths strategy

Why this matters

Forty-two PYQs, and a third of them also involve an ellipse: shared foci, or eccentricities with a given product. The relation b² = a²(e² − 1) and the constant difference of focal distances do most of the work. Four ideas cover the page.

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

A hyperbola

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

has the same list of features as an ellipse, with one sign changed:

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

, so

e>1

. The foci are still at

(\pm ae,0)

, the directrices at

x=\pm\frac{a}{e}

, and the latus rectum is still

\frac{2b^2}{a}

. Swapping the signs of the two squared terms gives the conjugate hyperbola, which opens up and down.

Definition

For $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ :

Transverse axis $2a$ , conjugate axis $2b$ .
$b^2=a^2(e^2-1)$ , $e>1$ .
Foci $(\pm ae,0)$ ; directrices $x=\pm\frac{a}{e}$ ; latus rectum $\frac{2b^2}{a}$ .
Opening up and down, $\frac{y^2}{b^2}-\frac{x^2}{a^2}=1$ : $a^2=b^2(e^2-1)$ , foci $(0,\pm be)$ , latus rectum $\frac{2a^2}{b}$ .
Conjugate hyperbolas $e$ , $e'$ : $\frac{1}{e^2}+\frac{1}{e'^2}=1$ .

The linking relation

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

Worked example

A hyperbola

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

has

e=\frac54

and latus rectum

\frac92

. Find

a

and

b

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 6 Apr 2026 Shift 1 · Q61Moderate

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

If the eccentricity e of the hyperbola

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

, passing through

(6,4\sqrt{3})

, satisfies

15\left( e^{2}+ 1 \right)= 34e

, then the length of the latus rectum of the hyperbola

\frac{x^{2}}{{\text{ }b}^{2}}-\frac{y^{2}}{2\left( a^{2}+ 1 \right)}= 1

is:

$e^2-1$ , not $1-e^2$

For a hyperbola

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

and

e>1

. A root

e<1

of a given equation belongs to an ellipse, not to this curve.

Drill 15 more on a, b, e, the latus rectum and the conjugate hyperbola

Concept 2 of 4: Ellipse and hyperbola together: shared foci and linked eccentricities

Many questions set a hyperbola beside an ellipse: they share foci, or one passes through the other's vertices, or their eccentricities multiply to

1

. The key is that the focal distance

c

comes from DIFFERENT formulas:

c^2=a^2-b^2

for the ellipse,

c^2=A^2+B^2

for the hyperbola. Find

c

from whichever curve is fully given, then use it on the other.

Definition

Ellipse: $c^2=a^2-b^2$ , $c=ae$ .
Hyperbola: $c^2=A^2+B^2$ , $c=Ae'$ .
Same foci: equal $c$ .
Through the other's foci or vertices: that fixes one semi-axis.
Given $ee'=1$ or a ratio: substitute one eccentricity into the other curve's relation.

The focal distance, two ways

c^2=a^2-b^2\ \text{(ellipse)},\qquad c^2=A^2+B^2\ \text{(hyperbola)}

Worked example

Find the hyperbola with the same foci as

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

and eccentricity

2

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 9 April 2024 · Q161Moderate

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

Let the foci of a hyperbola

H

coincide with the foci of the ellipse

E:\frac{(x- 1)^{2}}{100}+\frac{(y- 1)^{2}}{75}= 1

and the eccentricity of the hyperbola

H

be the reciprocal of the eccentricity of the ellipse E. If the length of the transverse axis of

H

\alpha

and the length of its conjugate axis is

\beta

, then

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

is equal to:

Do not reuse $a^2-b^2$ for the hyperbola

The ellipse gives

c^2=a^2-b^2

; the hyperbola needs

c^2=A^2+B^2

. Using the ellipse rule on the hyperbola gives a negative or wrong

B^2

Drill 12 more on ellipse and hyperbola together: shared foci and linked eccentricities

Concept 3 of 4: Focal distances: the constant difference, and foci anywhere

On a hyperbola the DIFFERENCE of the focal distances is constant:

|SP-S'P|=2a

. As with the ellipse, each is

e

times the distance to its directrix, giving

SP=ex-a

and

S'P=ex+a

on the right branch. When the foci are given as two points off the axes, the centre is their midpoint and

c

is half the distance between them; the eccentricity then gives the transverse semi-axis.

Definition

$|SP-S'P|=2a$ .
Right branch ( $x>0$ ), $S=(ae,0)$ : $SP=ex-a$ , $S'P=ex+a$ .
Product: $SP\cdot S'P=e^2x^2-a^2$ .
Triangle $PSS'$ : area $\frac12\cdot2ae\cdot|y|$ .
Foci given as points: centre = midpoint, $c$ = half their distance, $a=\frac{c}{e}$ .

Focal distances on the right branch

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

Worked example

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

, find the focal distances of the point with

x=8

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2026 · 22 Jan 2026 Shift 2 · Q61Moderate

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

Let

P(10,2\sqrt{15})

be a point on the hyperbola

\frac{x^{2}}{a^{2}}-\frac{y^{2}}{{\text{ }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

\bigtriangleup PSS^{'}

is equal to :

The DIFFERENCE is $2a$ , not the sum

For an ellipse

SP+S'P=2a

; for a hyperbola it is

|SP-S'P|=2a

. A question giving the SUM of focal distances on a hyperbola is giving

2ex

, not

2a

Drill 10 more on focal distances: the constant difference, and foci anywhere

Concept 4 of 4: The rectangular hyperbola, and hyperbolas that appear as loci

When

a=b

, the asymptotes are perpendicular and

e=\sqrt2

: this is a rectangular hyperbola. Turned through

45^\circ

it becomes

xy=c^2

, whose points are

\left(ct,\frac{c}{t}\right)

. Hyperbolas also turn up as loci: when two lines move with a parameter, multiply or combine their equations to remove it, and read the result.

Definition

$x^2-y^2=a^2$ : $e=\sqrt2$ , asymptotes $y=\pm x$ .
$xy=c^2$ : points $\left(ct,\frac{c}{t}\right)$ , asymptotes the axes, $e=\sqrt2$ .
$x^2y^2=1$ is the pair $xy=1$ and $xy=-1$ .
Loci: remove the parameter; if the result is $\frac{x^2}{A}-\frac{y^2}{B}=1$ , read $e=\sqrt{1+\frac{B}{A}}$ .

Rectangular hyperbola

xy=c^2:\ \left(ct,\frac{c}{t}\right),\qquad e=\sqrt2

Worked example

Find the eccentricity of the locus of

\left(2t,\frac{2}{t}\right)

Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2021 · Paper 3 · Q83Moderate

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

The point of intersection of the lines

\sqrt{3}\,kx + ky - 4\sqrt{3} = 0

and

\sqrt{3}\,x - y - 4\sqrt{3}\,k = 0

, as

k

varies, traces a conic whose eccentricity is

$xy=c^2$ has its axes along $y=\pm x$

The vertices of

xy=c^2

are

(c,c)

and

(-c,-c)

, not on the coordinate axes. Its transverse axis is the line

y=x

Drill 1 more on the rectangular hyperbola, and hyperbolas that appear as 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, the latus rectum and the conjugate hyperbola
The linking relation
$b^2=a^2(e^2-1),\qquad \text{LR}=\frac{2b^2}{a}$
Ellipse and hyperbola together: shared foci and linked eccentricities
The focal distance, two ways
$c^2=a^2-b^2\ \text{(ellipse)},\qquad c^2=A^2+B^2\ \text{(hyperbola)}$
Focal distances: the constant difference, and foci anywhere
Focal distances on the right branch
$SP=ex-a,\quad S'P=ex+a,\quad S'P-SP=2a$
The rectangular hyperbola, and hyperbolas that appear as loci
Rectangular hyperbola
$xy=c^2:\ \left(ct,\frac{c}{t}\right),\qquad e=\sqrt2$

Watch out for (4)

$e^2-1$ , not $1-e^2$
→ a, b, e, the latus rectum and the conjugate hyperbola
Do not reuse $a^2-b^2$ for the hyperbola
→ Ellipse and hyperbola together: shared foci and linked eccentricities
The DIFFERENCE is $2a$ , not the sum
→ Focal distances: the constant difference, and foci anywhere
$xy=c^2$ has its axes along $y=\pm x$
→ The rectangular hyperbola, and hyperbolas that appear as 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 42 questions