Question

Two points $A$ and $B$ on the curve $18x^2-9y^2=3$ such that slope of $CA\times$slope of $CB=-1$ where $C$ be the center of the curve, then value of $\frac{1}{C{A}^2}+\frac{1}{C{B}^2}$ is

Answer 1

Let $CA=a;CB=b$
curve $18x^2-9y^2=3$ is hyperbola
slopes of $CA\times CB=-1$
$\therefore$ lines $CA$ and $CB$ are perpendicular to each other
Coordinates of $A\equiv (a\cos \theta ,a\sin \theta )$ and
$B\equiv (b\cos (\frac{\pi }{2}+\theta ),b\sin (\frac{\pi }{2}+\theta ))\equiv (-b\sin \theta ,b\cos \theta )$
$\because A\text{and}B$ lies on the hyperbola also
$\therefore 6a^2{\cos }^2\theta -3a^2{\sin }^2\theta =1\Rightarrow 6{\cos }^2\theta -3{\sin }^2\theta =\frac{1}{a^2}...\left(1\right)$
and
$6b^2{\sin }^2\theta -3b^2{\cos }^2\theta =1\Rightarrow 6{\sin }^2\theta -3{\cos }^2\theta =\frac{1}{b^2}...\left(2\right)$
From equation $\left(1\right)+\left(2\right)$
$\frac{1}{a^2}+\frac{1}{b^2}=3$