The locus of the midpoints of the portion of | vedclass

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(C) Given the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$,where $a^2=2$ and $b^2=1$.The equation of the tangent at any point $(\sqrt{2}\cos 	heta, \sin \

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$\frac{1}{2x^2}+\frac{1}{4y^2}=1$

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(C) Given the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$,where $a^2=2$ and $b^2=1$.The equation of the tangent at any point $(\sqrt{2}\cos 	heta, \sin 	heta)$ is given by $\frac{x \cos 	heta}{\sqrt{2}} + y \sin 	heta = 1$.The tangent intersects the $x$-axis at point $A$ where $y=0$,so $A = (\frac{\sqrt{2}}{\cos 	heta}, 0)$.The tangent intersects the $y$-axis at point $B$ where $x=0$,so $B = (0, \frac{1}{\sin 	heta})$.Let $(h, k)$ be the midpoint of the segment $AB$. Then $h = \frac{\sqrt{2}}{2 \cos 	heta}$ and $k = \frac{1}{2 \sin 	heta}$.From these,we have $\cos 	heta = \frac{\sqrt{2}}{2h} = \frac{1}{\sqrt{2}h}$ and $\sin 	heta = \frac{1}{2k}$.Using the identity $\sin^2 	heta + \cos^2 	heta = 1$,we get $(\frac{1}{2k})^2 + (\frac{1}{\sqrt{2}h})^2 = 1$.This simplifies to $\frac{1}{4k^2} + \frac{1}{2h^2} = 1$.Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

The locus of the midpoints of the portion of the tangents of the ellipse $\frac{x^2}{2}+\frac{y^2}{1}=1$ intercepted between the coordinate axes is

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