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 $a^2 b^2=\alpha \sqrt{2}-\beta, \alpha, \beta \in \mathbb{N}$.

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Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}_1 \equiv(-\mathrm{ae}, 0) \equiv \mathrm{P}(-3,0) \\ & \Rightarrow \mathrm{ae}=3\end{aligned}$

$\begin{aligned} & \tan 45^{\circ}=\frac{b^2 / a}{2 a e} \\ & 2 a e=\frac{b^2}{a} \\ & b^2=6 a \\ & \text { Also } a^2 e^2=a^2+b^2 \\ & 9=a^2+6 a \\ & a^2+6 a-9=0 \\ & a=-3 \pm 3 \sqrt{2}=-3(1 \pm \sqrt{2}) \\ & \therefore a^2 b^2=a^2 \cdot 6 a=6 a^3 \\ & =6(135 \sqrt{2}-189) \\ & \alpha=810 \text { and } \beta=1134 \\ & \therefore \alpha+\beta=1944\end{aligned}$

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…

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