Let \(\mathrm{H}: \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1\) be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is \(\frac{8}{3}\). If the line \(\mathrm{x}=\alpha\) intersects the hyperbola H at the points A and B such that the area of the triangle AOB is \(4 \sqrt{15}\), where O is the origin, then \(\alpha^{2}\) equals

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Hyperbola

Let

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

be a hyperbola such that the distance between its foci is 6 and the distance between its directrices is

\frac{8}{3}

. If the line

x = α

intersects the hyperbola H at the points A and B such that the area of the triangle AOB is

415

, where O is the origin, then

α^{2}

equals

Answer: (B)

Distance between foci

2 a e = 6

gives

a e = 3

; distance between directrices

\frac{2 a}{e} = \frac{8}{3}

gives

\frac{a}{e} = \frac{4}{3}

. Multiplying,

a^{2} = 4

, and dividing,

e^{2} = \frac{9}{4}

e = \frac{3}{2}

. Then

b^{2} = a^{2} (e^{2} - 1) = 4 (\frac{9}{4} - 1) = 5

, and

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

. On

x = α

y^{2} = \frac{5 ( α ^{2} - 4 )}{4}

, so

A, B = (α, \pm \frac{5 ( α ^{2} - 4 )}{4})

. Triangle

A O B

has base

∣ A B ∣ = 2∣ y ∣

and height

∣ α ∣

, giving area

∣ α ∣ \frac{5 ( α ^{2} - 4 )}{4} = 415

. Squaring,

\frac{5 α ^{2} ( α ^{2} - 4 )}{4} = 240

, so

α^{4} - 4 α^{2} - 192 = 0

, i.e.

(α^{2} - 16) (α^{2} + 12) = 0

. Taking the real value,

α^{2} = 16

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Single correct option