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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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$\sqrt{3 / 2}$

$1 / 2 \sqrt{2}$

$1 / \sqrt{2}$

$1 / 2$

answer is C.

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Detailed Solution

$\frac{x^{2}}{4} + \frac{y^{2}}{2} = 1 \to Q (2 \cos θ, \sqrt{2} \sin θ)$

$R = (x, y) = (\frac{2 \cos θ + 4}{2}, \frac{\sqrt{2} \sin θ + 3}{2})$

Eliminate $' θ' : \frac{{(x - 2)}^{2}}{1} + \frac{{(y - 3 / 2)}^{2}}{(1 / 2)} = 1$

$e = \sqrt{1 - 1 / 2} = \frac{1}{\sqrt{2}}$

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