JEE Main 2026 — Apr 4, Shift 2 — Mathematics solution

JEE Main2026Apr 4, Shift 2Mathematics

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 $A O B$ is $415$ , where $O$ is the origin, then $α^{2}$ equals

A
12
B
16
C
24
D
25

Solution

Given: $2 a e = 6 \Rightarrow a e = 3$ , and $\frac{2 a}{e} = \frac{8}{3} \Rightarrow \frac{a}{e} = \frac{4}{3}$ .

Multiplying: $a^{2} = a e \cdot \frac{a}{e} = 3 \cdot \frac{4}{3} = 4$ . Dividing: $e^{2} = \frac{a e}{a / e} = \frac{3}{4/3} = \frac{9}{4} \Rightarrow e = \frac{3}{2}$ .

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

Hyperbola: $\frac{x ^{2}}{4} - \frac{y ^{2}}{5} = 1$ . On $x = α$ : $y^{2} = \frac{5 ( α ^{2} - 4 )}{4}$ , so $A, B = (α, \pm \frac{5 ( α ^{2} - 4 )}{4})$ .

Area of $△ A O B = \frac{1}{2} \cdot ∣ α ∣ \cdot 2 \frac{5 ( α ^{2} - 4 )}{4} = ∣ α ∣ \frac{5 ( α ^{2} - 4 )}{4} = 415$ .

Squaring: $α^{2} \cdot \frac{5 ( α ^{2} - 4 )}{4} = 240 \Rightarrow α^{2} (α^{2} - 4) = 192$ .

Let $u = α^{2}$ : $u^{2} - 4 u - 192 = 0 \Rightarrow u = \frac{4 + 16 + 768}{2} = \frac{4 + 28}{2} = 16$ .

$∴ α^{2} = 16$ .

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