Mathematics · Hyperbola

JEE Main 2025 — 7 April, Morning Shift — Question 44

Consider the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ having one of its focus at $P(-3,0)$ . If the latus rectum through its

other focus subtends a right angle at $P$ and $a^{2} b^{2}=\alpha \sqrt{2}-\beta, \alpha, \beta \in \mathbb{N}$ , then $\alpha+\beta$ is $\_\_\_\_$ .

Answer: 1944

Numerical answer — enter this value.

ae = 3,\qquad \tan 45^\circ = 1

\tan 45^\circ = ae \cdot \frac{b^2}{6}

1 = 3 \cdot \frac{b^2}{6} \Rightarrow \frac{b^2}{2} = 1 \Rightarrow b^2 = 2

\tan 45^\circ = ae \cdot \frac{b^2}{6} \Rightarrow \frac{b^2}{a} = 6 \quad \text{(i)}

a\sqrt{1+\frac{b^2}{a^2}} = 3

\frac{a\sqrt{a^2+b^2}}{a} = 3 \Rightarrow \sqrt{a^2+b^2} = 3

a^2+b^2 = 9 \quad \text{(ii)}

From (i) and (ii),

b^2 = 6a

a^2 + 6a = 9 \Rightarrow a^2 + 6a - 9 = 0

a = \frac{-6 + \sqrt{36 + 36}}{2} = \frac{-6 + 6\sqrt{2}}{2} = 3(\sqrt{2}-1)

b^2 = 6a = 18(\sqrt{2}-1)

a^2 b^2 = \left[3(\sqrt{2}-1)\right]^2 \cdot 18(\sqrt{2}-1)

= 162(5\sqrt{2}-7)

Let

a^2 b^2 = \alpha\sqrt{2} - \beta

\alpha = 162 \times 5,\qquad \beta = 162 \times 7

\alpha + \beta = 162(5+7) = \boxed{1944}

Answer key and solution verified before publishing.

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