If P(theta) and Qleft(frac{pi}{2}+thetaright) | vedclass

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(D) The coordinates of points $P$ and $Q$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $P = (a \cos 	heta, b \sin 	heta)$ and $Q = (a \cos

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$\sqrt{2}$

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(D) The coordinates of points $P$ and $Q$ on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $P = (a \cos 	heta, b \sin 	heta)$ and $Q = (a \cos(\frac{\pi}{2}+	heta), b \sin(\frac{\pi}{2}+	heta)) = (-a \sin 	heta, b \cos 	heta)$.Let the midpoint of $PQ$ be $(x, y)$. Then,$x = \frac{a(\cos 	heta - \sin 	heta)}{2} \Rightarrow \frac{2x}{a} = \cos 	heta - \sin 	heta$$y = \frac{b(\sin 	heta + \cos 	heta)}{2} \Rightarrow \frac{2y}{b} = \sin 	heta + \cos 	heta$Squaring and adding these equations:$(\frac{2x}{a})^2 + (\frac{2y}{b})^2 = (\cos 	heta - \sin 	heta)^2 + (\sin 	heta + \cos 	heta)^2$$\frac{4x^2}{a^2} + \frac{4y^2}{b^2} = (\cos^2 	heta + \sin^2 	heta - 2 \sin 	heta \cos 	heta) + (\sin^2 	heta + \cos^2 	heta + 2 \sin 	heta \cos 	heta) = 2$$\frac{x^2}{a^2/2} + \frac{y^2}{b^2/2} = 1$Comparing this with $\frac{x^2}{\alpha^2} + \frac{y^2}{\beta^2} = 1$,we get $\alpha = \frac{a}{\sqrt{2}}$ and $\beta = \frac{b}{\sqrt{2}}$.Therefore,$\frac{a+b}{\alpha+\beta} = \frac{a+b}{\frac{a}{\sqrt{2}} + \frac{b}{\sqrt{2}}} = \frac{a+b}{\frac{1}{\sqrt{2}}(a+b)} = \sqrt{2}$.

If $P(\theta)$ and $Q\left(\frac{\pi}{2}+\theta\right)$ are two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the locus of the midpoint of $PQ$ is $\frac{x^2}{\alpha^2}+\frac{y^2}{\beta^2}=1$,then $\frac{a+b}{\alpha+\beta}=$

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