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

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

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

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

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

answer is C.

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Detailed Solution

$eq of Tangent is \frac{x}{\sqrt{2}} \cos + \frac{y}{1} sinθ = 1 meets coordinate axes at A, B$

$A = (\frac{\sqrt{2}}{cosθ}, 0) B = (0, \frac{1}{2 sinθ}) mid point of AB = P (x, y) P = (\frac{\sqrt{2}}{2 cosθ}, \frac{1}{2 sinθ}) cosθ = \frac{1}{\sqrt{2} x} sinθ = \frac{1}{2 y} {(\frac{1}{\sqrt{2} x})}^{2} + {(\frac{1}{2 y})}^{2} = 1$

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