Let S equiv frac{x^2}{a^2}+frac{y^2}{b^2}-1=0 | vedclass

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(D) Given the ellipses $S^{\prime} \equiv \frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1$. Points $P(a \cos 	heta, b \sin 	heta)$ and $Q\left(a \cos \le

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$\alpha^2 \beta^2$

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(D) Given the ellipses $S^{\prime} \equiv \frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1$. Points $P(a \cos 	heta, b \sin 	heta)$ and $Q\left(a \cos \left(\frac{\pi}{2}+	hetaight), b \sin \left(\frac{\pi}{2}+	hetaight)ight)$ lie on $S^{\prime}$. Simplifying $Q$,we get $Q \equiv (-a \sin 	heta, b \cos 	heta)$. Since $P$ lies on $S^{\prime}$,we have: $\frac{a^2 \cos^2 	heta}{\alpha^2} + \frac{b^2 \sin^2 	heta}{\beta^2} = 1$ $(i)$ Since $Q$ lies on $S^{\prime}$,we have: $\frac{a^2 \sin^2 	heta}{\alpha^2} + \frac{b^2 \cos^2 	heta}{\beta^2} = 1$ $(ii)$ Adding $(i)$ and $(ii)$: $\frac{a^2}{\alpha^2}(\cos^2 	heta + \sin^2 	heta) + \frac{b^2}{\beta^2}(\sin^2 	heta + \cos^2 	heta) = 1 + 1$ $\frac{a^2}{\alpha^2} + \frac{b^2}{\beta^2} = 2$ $\frac{a^2 \beta^2 + b^2 \alpha^2}{\alpha^2 \beta^2} = 2$ Therefore,$\frac{1}{2}(a^2 \beta^2 + b^2 \alpha^2) = \alpha^2 \beta^2$.

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