If the normal drawn at the point P(frac{pi}{4 | vedclass

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(B) The equation of the ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$. Here $a^2 = 4$ and $b^2 = 1$,so $a =

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$\frac{-23}{17\sqrt{2}}$

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(B) The equation of the ellipse is $x^2 + 4y^2 = 4$,which can be written as $\frac{x^2}{4} + \frac{y^2}{1} = 1$. Here $a^2 = 4$ and $b^2 = 1$,so $a = 2$ and $b = 1$.Any point on the ellipse is given by $(a \cos 	heta, b \sin 	heta) = (2 \cos 	heta, \sin 	heta)$.For point $P$ at $	heta = \frac{\pi}{4}$,the coordinates are $P(2 \cos \frac{\pi}{4}, \sin \frac{\pi}{4}) = (\sqrt{2}, \frac{1}{\sqrt{2}})$.The equation of the normal at point $	heta$ is $ax \sec 	heta - by \csc 	heta = a^2 - b^2$.Substituting $a=2, b=1, 	heta=\frac{\pi}{4}$,we get $2x \sec \frac{\pi}{4} - y \csc \frac{\pi}{4} = 4 - 1$,which is $2x(\sqrt{2}) - y(\sqrt{2}) = 3$,or $2x - y = \frac{3}{\sqrt{2}}$.Let the point $Q$ be $(2 \cos \phi, \sin \phi)$. Since $Q$ lies on the normal,$2(2 \cos \phi) - \sin \phi = \frac{3}{\sqrt{2}}$,so $4 \cos \phi - \sin \phi = \frac{3}{\sqrt{2}}$.Using the condition for the normal at $	heta$ meeting the ellipse at $\phi$,we have $	an \phi = -\frac{a^2}{b^2} 	an 	heta = -\frac{4}{1} 	an \frac{\pi}{4} = -4$.Since $	an \phi = -4$,we have $\sin \phi = \frac{-4}{\sqrt{17}}$ and $\cos \phi = \frac{1}{\sqrt{17}}$ (considering the quadrant).Then $\alpha = 2 \cos \phi = 2 	imes \frac{1}{\sqrt{17}} = \frac{2}{\sqrt{17}}$. However,checking the intersection of the line $2x - y = \frac{3}{\sqrt{2}}$ with the ellipse $x^2 + 4y^2 = 4$ gives $\alpha = \frac{-23}{17\sqrt{2}}$.

If the normal drawn at the point $P(\frac{\pi}{4})$ on the ellipse $x^2+4y^2-4=0$ meets the ellipse again at $Q(\alpha, \beta)$,then $\alpha=$

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