Question

If C is the center and A,B are two points on the conic ${4x}^2$ + ${9y}^2$ -8$x$ -36y +4=0
$\angle$ ACB =$\frac{\pi }{2}$ , then ${CA}^{-2}$ + ${CB}^{-2}$ + $\frac{23}{36}$ =

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Answer 1

Given eqn. is 4( $x-1{)}^2$ + 9( $y-2{)}^2$ =36 which is an ellipse C=(1,2)
Let CA make an angle $\theta$ with the major axis.
A= (1 + CA cos$\theta$ , 2 + sin$\theta$)
B= (1+ CB cos [$\frac{\pi }{2}$ +$\theta$] , 2 + CB sin[$\frac{\pi }{2}$ +$\theta$])
A,B then on the ellipse
$\Rightarrow$ ${CA}^2$ ( 4 ${cos}^2$$\theta$ +9${sin}^2$$\theta$) =36
${CB}^2$ ( 4 ${sin}^2$$\theta$ +9${cos}^2$$\theta$) =36
$\Rightarrow$ ${CA}^{-2}$ + ${CB}^{-2}$ = $\frac{13}{36}$