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Question

Mathematics Question on Ellipse

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:

A

32\frac{\sqrt3}{2}

B

122\frac{1}{2\sqrt2}

C

12\frac{1}{\sqrt2}

D

12\frac{1}{2}

Answer

12\frac{1}{\sqrt2}

Explanation

Solution

The correct answer is (C) : 12\frac{1}{\sqrt2}
Let P(2cosθ,√2sinθ)P(2cosθ, √2sinθ ) be any point on ellipse x24+y22=1\frac{x²}{4} + \frac{y²}{2} = 1
and Q(4,3) and let (h, k) be the mid point of PQ then
h = \frac{2cosθ+4}{2}$$,k = \frac{\sqrt2Sinθ+3}{2}
∴cosθ=h−2,sinθ=2k−32∴ cosθ = h - 2 , sinθ = \frac{2k - 3}{\sqrt2}
∴(h−2)2+(2k−32)2=1∴ ( h - 2 )² + ( \frac{2k - 3}{\sqrt2} )^2 = 1
⇒(x−2)21+(y−32)212=1⇒ \frac{( x - 2 )²} {1} + \frac{( \frac{y - 3}{2} )^2 } {\frac{1}{2}} = 1
∴e=1−12=12∴ e = \sqrt{1 - \frac{1}{2}} = \frac{1}{\sqrt2}