The locus of the mid point of the line segmen | vedclass

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Question

$\frac{ x ^{2}}{4}+\frac{ y ^{2}}{2}=1$

Coordinate of $D$ is

$\left(\frac{2 \cos 	heta+4}{2}, \frac{\sqrt{2} \sin 	heta+3}{2}ight) \equiv(h,

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

$\frac{ x ^{2}}{4}+\frac{ y ^{2}}{2}=1$

Coordinate of $D$ is

$\left(\frac{2 \cos 	heta+4}{2}, \frac{\sqrt{2} \sin 	heta+3}{2}ight) \equiv(h, k)$

$\frac{2 h -4}{2}=\cos 	heta$

$\frac{2 k -3}{\sqrt{2}}=\sin 	heta$

$(i)^{2}+(	ext { ii })^{2}$, then we get

$\left(\frac{2 h -4}{2}ight)^{2}+\left(\frac{2 k -3}{\sqrt{2}}ight)^{2}=1 \Rightarrow \frac{( x -2)^{2}}{1}+\frac{\left( y -\frac{3}{2}ight)^{2}}{\left(\frac{1}{2}ight)}=1$

$	herefore \quad$ Required eccentricity is

$e =\sqrt{1-\frac{1}{2}}=\frac{1}{\sqrt{2}}$

The locus of the mid point of the line segment joining the point $(4,3)$ and the points on the ellipse $x^{2}+2 y^{2}=4$ is an ellipse with eccentricity

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