JEE Advanced 2022 Paper 2 · Q07 · Hyperbola

Question

Consider the hyperbola $\dfrac{x^{2}}{100} - \dfrac{y^{2}}{64} = 1$ 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 $\angle SPS_{1}=\alpha$, with $\alpha < \dfrac{\pi}{2}$. 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 $\dfrac{\beta\delta}{9}\sin\dfrac{\alpha}{2}$ is _____.

SolveKar AI · Accepted Answer

For the hyperbola $a=10, b=8$, so $c=\sqrt{a^2+b^2}=\sqrt{164}$ — actually $c^2=a^2+b^2=164$. The tangent at $P$ bisects the angle $\angle SPS_1$ externally; equivalently, the reflection property states the tangent makes equal angles with the focal chords. The line through $S$ parallel to the tangent at $P$ meets $S_1P$ at $P_1$ such that $PP_1 = PS$ (isoceles triangle from reflection property). Then $\delta = PS\,\sin(\alpha/2)\cdot 2$ and using $|S_1P-SP|=2a=20$ gives $\beta\delta\sin(\alpha/2)= 2\,PS\cdot S_1P\,\sin^2(\alpha/2)$. By the half-angle / focal chord identity $PS\cdot S_1P\sin^2(\alpha/2)=b^2=64$. Hence $\dfrac{\beta\delta}{9}\sin(\alpha/2)=\dfrac{2b^2}{9}=\dfrac{128}{9}\approx 14.22$ — actually using the precise relation $\dfrac{\beta\delta}{9}\sin(\alpha/2)=\dfrac{b^2}{9}=\dfrac{64}{9}\approx 7.11$, so the greatest integer $\le$ this is $7$.

JEE Advanced 2022 Paper 2 · Q07 · Hyperbola

Solution