JEE Main 2025 (Online) 4th April Evening ShiftQuestion 1
Let the sum of the focal distances of the point $P(4,3)$ on the hyperbola $H:\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ be $8\sqrt{\frac{5}{3}}$. If for $H$, the length of the latus rectum is $l$ and the product of the focal distances of the point $P$ is $m$, then $9l^{2}+6m$ is equal to :
Hyperbola
Solution
For $x=4>0$ the focal distances are $ex\pm a$; their sum is $2ex=8e=8\sqrt{\tfrac53}$, so $e^2=\tfrac53$, $b^2=\tfrac23a^2$. $P(4,3)$ on $H$: $\dfrac{16}{a^2}-\dfrac{27}{2a^2}=1\Rightarrow a^2=\dfrac52$, $b^2=\dfrac53$. $l^2=\dfrac{4b^4}{a^2}=\dfrac{40}{9}$; $m=e^2x^2-a^2=\dfrac{80}{3}-\dfrac52=\dfrac{145}{6}$. $9l^2+6m=40+145=185$.