$C$ is the centre of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ and $A$ and $B$ are two po

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$C$ is the centre of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{9} = 1$ and $A$ and $B$ are two points on the ellipse such $\angle ACD = 90^{\circ}$. Then $\frac{1}{CA^{2}} + \frac{1}{CB^{2}} =$ :

Accepted Answer

Sol.(a) Put ( $rcos	heta,rsin	heta$ ) to the equation of ellipse$${\Rightarrow r^{2}\left( \frac{\cos^{2}	heta}{a^{2}} + \frac{\sin^{2}	heta}{b^{2}} ight) = 1 }{\Rightarrow \frac{1}{r_{1}^{2}} + \frac{1}{r_{2}^{2}} = \left( \frac{\cos^{2}	heta}{a^{2}} + \frac{\sin^{2}	heta}{b^{2}} ight) + \left( \frac{\cos^{2}\left( 	heta + \frac{\pi}{2} ight)}{a^{2}} + \frac{\sin^{2}\left( 	heta + \frac{\pi}{2} ight)}{b^{2}} ight)}$$$$= \frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{16} + \frac{1}{9} = \frac{25}{144}$$

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