Question

Consider the hyperbola $\frac{x^2}{100}-\frac{y^2}{64}=1$ with foci at $S$ and $S_1$, where $S$ lies on the positive $x$-axis Let $P$ be a point on the hyperbola, in the first quadrant Let $\angle \operatorname{SPS}_1=\alpha$, with $\alpha<\frac{\pi}{2}$. The straight line passing through the point $S$ and having the same slope as that of the tangent at $P$ to the hyperbola, intersects the straight line $S _1 P$ at $P _1$ Let $\delta$ be the distance of $P$ from the straight line $SP _1$, and $\beta= S _1 P$. Then the greatest integer less than or equal to $\frac{\beta \delta}{9} \sin \frac{\alpha}{2}$ is _____.

Answer 1

The answer is 7.

Answer 2

$\begin{aligned} &Given\\& S P-P P=20 \\ & We\ have\ \beta-\frac{\delta}{\sin \frac{\alpha}{2}}=20 \\ &Squaring\ the\ above\ equation\ and\ rearranging\ terms,\ we\ obtain  \\ & \beta^2+\frac{\delta^2}{\sin ^2 \frac{\alpha}{2}}-400=\frac{2 \beta \delta}{\sin \frac{\alpha}{2}} \\ &The\ reciprocal\ of\ SP\ is\ related\ to\ \delta\ as\ \frac{1}{S P}=\frac{\sin \frac{\alpha}{2}}{\delta} \\ & \cos \alpha=\frac{S P^2+\beta^2-656}{2 \beta \frac{\theta}{\sin \frac{\sigma}{2}}} \\ & =\frac{\frac{2 \beta \delta}{\frac{\sin }{2}}-256}{\frac{2 \beta S}{\sin \frac{\alpha}{2}}}=\cos \alpha \\ & Solving\ for\ \lambda\\ & \frac{\lambda-128}{\lambda}=\cos \alpha \\ & \lambda(1-\cos \alpha)=128 \\ & \frac{\beta \delta}{\sin \frac{\alpha}{2}} \cdot 2 \sin ^2 \frac{\alpha}{2}=128 \\ & \frac{\beta \delta}{9} \sin \frac{\alpha}{2}=\frac{64}{9}\\ &\Rightarrow\left[\frac{\beta \delta}{9} \sin \frac{\alpha}{2}\right]=7 \text { where [.] denotes greatest integer function } \\ & \end{aligned}$

So, the answer is 7.