JEE Advanced 2022 · Paper 2, Mathematics question 7

Here $a=10$, $b=8$. Write $r=SP$ and $\beta=S_1P$, so $\beta-r=2a=20$.

Isosceles triangle. The tangent at $P$ bisects $\angle SPS_1$. The line $SP_1$ is parallel to it, so in triangle $PSP_1$ the angles at $S$ and $P_1$ both equal $\frac\alpha2$. Hence $PS=PP_1$, and the distance from $P$ to $SP_1$ is
$$\delta=r\sin\frac\alpha2.$$

Focal triangle. By the cosine rule in $SPS_1$, with $SS_1=2c$:
$$4c^2=r^2+\beta^2-2r\beta\cos\alpha=(\beta-r)^2+4r\beta\sin^2\frac\alpha2.$$
So $r\beta\sin^2\frac\alpha2=c^2-a^2=b^2=64$.

Answer.
$$\frac{\beta\delta}9\sin\frac\alpha2=\frac{r\beta\sin^2\frac\alpha2}9=\frac{64}9\approx7.1,$$
so the greatest integer is 7.