JEE MAIN - Mathematics (2025 - 7th April Morning Shift - No. 25)

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

Explanation

JEE Main 2025 (Online) 7th April Morning Shift Mathematics - Hyperbola Question 4 English Explanation

$$\begin{aligned} & a e=3, \tan 45^{\circ}=a e=3 \frac{\frac{b^2}{a}}{6} \Rightarrow \frac{b^2}{a}=6 \ldots \text { (i) } \quad\text{..... (i)}\\ & a \sqrt{\left(1+\frac{b^2}{a^2}\right)}=3 \\ & a^2+b^2=9 \quad\text{..... (ii)}\\ & \Rightarrow a^2-6 a+9=0 \Rightarrow a=3(\sqrt{2}-1) \\ & \Rightarrow a^2 b^2=9(3-2 \sqrt{2}) \cdot 6 \cdot 3(\sqrt{2}-1) \\ & \quad=162(5 \sqrt{2}-7) \\ & \Rightarrow \alpha=162 \times 5, \beta=162 \times 7 \\ & \Rightarrow \alpha+\beta=162 \times 12=1944 \end{aligned}$$

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