The locus of the midpoint of the line segment | vedclass

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(C) Let $P = (4, 3)$ and $Q = (2\cos	heta, \sqrt{2}\sin	heta)$ be a point on the ellipse $x^{2} + 2y^{2} = 4$.Let $D(h, k)$ be the midpoint of $PQ$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) Let $P = (4, 3)$ and $Q = (2\cos	heta, \sqrt{2}\sin	heta)$ be a point on the ellipse $x^{2} + 2y^{2} = 4$.Let $D(h, k)$ be the midpoint of $PQ$.Then,$h = \frac{4 + 2\cos	heta}{2} = 2 + \cos	heta \implies \cos	heta = h - 2$.And $k = \frac{3 + \sqrt{2}\sin	heta}{2} \implies \sqrt{2}\sin	heta = 2k - 3 \implies \sin	heta = \frac{2k - 3}{\sqrt{2}}$.Using the identity $\cos^{2}	heta + \sin^{2}	heta = 1$,we get:$(h - 2)^{2} + \left(\frac{2k - 3}{\sqrt{2}}ight)^{2} = 1$$(h - 2)^{2} + \frac{4(k - 1.5)^{2}}{2} = 1$$(h - 2)^{2} + 2(k - 1.5)^{2} = 1$Dividing by $1$,we get $\frac{(h - 2)^{2}}{1} + \frac{(k - 1.5)^{2}}{1/2} = 1$.This is an ellipse with $a^{2} = 1$ and $b^{2} = 1/2$.Since $a^{2} > b^{2}$,the eccentricity $e$ is given by $e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{1/2}{1}} = \sqrt{1 - 1/2} = \sqrt{1/2} = \frac{1}{\sqrt{2}}$.

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

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