HyperbolamediumPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Hyperbola Triangle Area 4√15: α² = 16 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question

Let

H:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1

be a hyperbola such that the distance between its foci is

6

and the distance between its directrices is

\dfrac83

. If the line

x=\alpha

intersects

H

A

and

B

such that the area of

\triangle AOB

4\sqrt{15}

, where

O

is the origin, then

\alpha^2

equals

12

16

correct

24

25

Solution

Step 1: Distance between foci

2ae=6\Rightarrow ae=3

. Distance between directrices

\dfrac{2a}{e}=\dfrac83\Rightarrow\dfrac{a}{e}=\dfrac43

. Step 2: Multiply:

a^2=3\cdot\dfrac43=4\Rightarrow a=2

; divide:

e^2=\dfrac{3}{4/3}=\dfrac94\Rightarrow e=\dfrac32

. Then

b^2=a^2(e^2-1)=4\left(\dfrac94-1\right)=5

. So

H:\dfrac{x^2}{4}-\dfrac{y^2}{5}=1

. Step 3: At

x=\alpha

y=\pm\sqrt{\dfrac{5(\alpha^2-4)}{4}}

, so

A\left(\alpha,\sqrt{\tfrac{5(\alpha^2-4)}{4}}\right)

B\left(\alpha,-\sqrt{\tfrac{5(\alpha^2-4)}{4}}\right)

. Step 4: Area of

\triangle AOB=\dfrac12\cdot|\alpha|\cdot AB=\dfrac12|\alpha|\sqrt{5(\alpha^2-4)}=4\sqrt{15}

. Step 5: Square:

\dfrac14\alpha^2\cdot5(\alpha^2-4)=16\cdot15=240\Rightarrow\alpha^2(\alpha^2-4)=192

. Put

u=\alpha^2

u^2-4u-192=0\Rightarrow(u-16)(u+12)=0\Rightarrow u=16

. So

\alpha^2=16

. Correct answer: (2)

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