If tangents are drawn to the ellipse x^2 + 2y | vedclass

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

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

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(C) The given ellipse is $x^2 + 2y^2 = 2$,which can be written as $\frac{x^2}{2} + \frac{y^2}{1} = 1$. Here $a^2 = 2$ and $b^2 = 1$.The equation of the tangent to the ellipse at point $(a \cos 	heta, b \sin 	heta)$ is $\frac{x}{a} \cos 	heta + \frac{y}{b} \sin 	heta = 1$.The tangent intersects the $x$-axis at $A \left( \frac{a}{\cos 	heta}, 0 ight)$ and the $y$-axis at $B \left( 0, \frac{b}{\sin 	heta} ight)$.Let $P(h, k)$ be the midpoint of the segment $AB$. Then:$h = \frac{a}{2 \cos 	heta} \Rightarrow \cos 	heta = \frac{a}{2h}$$k = \frac{b}{2 \sin 	heta} \Rightarrow \sin 	heta = \frac{b}{2k}$Using the identity $\cos^2 	heta + \sin^2 	heta = 1$,we get:$\left( \frac{a}{2h} ight)^2 + \left( \frac{b}{2k} ight)^2 = 1$$\frac{a^2}{4h^2} + \frac{b^2}{4k^2} = 1$Substituting $a^2 = 2$ and $b^2 = 1$:$\frac{2}{4h^2} + \frac{1}{4k^2} = 1$$\frac{1}{2h^2} + \frac{1}{4k^2} = 1$Replacing $(h, k)$ with $(x, y)$,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

If tangents are drawn to the ellipse $x^2 + 2y^2 = 2$ at all points on the ellipse other than its four vertices,then the midpoints of the tangents intercepted between the coordinate axes lie on the curve:

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