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

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(C) The equation of the ellipse is $x^2+2y^2=2$,which can be written as $\frac{x^2}{2}+\frac{y^2}{1}=1$.Any point $P$ on the ellipse is $(\sqrt{2}\cos

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The equation of the ellipse is $x^2+2y^2=2$,which can be written as $\frac{x^2}{2}+\frac{y^2}{1}=1$.Any point $P$ on the ellipse is $(\sqrt{2}\cos	heta, \sin	heta)$.The equation of the tangent at $P$ is $\frac{x(\sqrt{2}\cos	heta)}{2} + \frac{y(\sin	heta)}{1} = 1$,which simplifies to $\frac{x\cos	heta}{\sqrt{2}} + y\sin	heta = 1$.The intercepts made by the tangent on the coordinate axes are $A\left(\frac{\sqrt{2}}{\cos	heta}, 0ight)$ and $B\left(0, \frac{1}{\sin	heta}ight)$.Let $(h, k)$ be the mid-point of $AB$. Then $h = \frac{\sqrt{2}}{2\cos	heta}$ and $k = \frac{1}{2\sin	heta}$.This implies $\cos	heta = \frac{1}{\sqrt{2}h}$ and $\sin	heta = \frac{1}{2k}$.Using the identity $\cos^2	heta + \sin^2	heta = 1$,we get $\left(\frac{1}{\sqrt{2}h}ight)^2 + \left(\frac{1}{2k}ight)^2 = 1$.Thus,the locus is $\frac{1}{2x^2} + \frac{1}{4y^2} = 1$.

If tangents are drawn to the ellipse $x^2+2y^2=2$,then the locus of the mid-points of the intercepts made by those tangents between the coordinate axes is

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