Let C be the centre of an ellipse and PQ be a | vedclass

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(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let $P = (a \cos 	heta, b \sin 	heta)$ and $Q = (a \cos \phi, b \sin

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an ellipse

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(C) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let $P = (a \cos 	heta, b \sin 	heta)$ and $Q = (a \cos \phi, b \sin \phi)$. Since $\angle PCQ = 90^{\circ}$,the product of the slopes of $CP$ and $CQ$ is $-1$. Thus,$\frac{b \sin 	heta}{a \cos 	heta} 	imes \frac{b \sin \phi}{a \cos \phi} = -1$,which implies $\frac{b^2}{a^2} 	an 	heta 	an \phi = -1$,or $	an 	heta 	an \phi = -\frac{a^2}{b^2}$. The intersection point $R(h, k)$ of the tangents at $P$ and $Q$ is given by $h = a \frac{\cos(\frac{	heta+\phi}{2})}{\cos(\frac{	heta-\phi}{2})}$ and $k = b \frac{\sin(\frac{	heta+\phi}{2})}{\cos(\frac{	heta-\phi}{2})}$. Squaring and rearranging,we find that $\frac{h^2}{a^4} + \frac{k^2}{b^4} = \frac{1}{a^2 \cos^2(\frac{	heta-\phi}{2})} + \frac{1}{b^2 \cos^2(\frac{	heta-\phi}{2})}$. Using the condition $	an 	heta 	an \phi = -\frac{a^2}{b^2}$,it can be shown that the locus of $R$ is $\frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{a^2} + \frac{1}{b^2}$,which represents another ellipse.

Let $C$ be the centre of an ellipse and $PQ$ be a chord of it with $\angle PCQ = 90^{\circ}$. If $R$ is the point of intersection of the tangents to the ellipse at $P$ and $Q$,then $R$ lies on

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