Given,
4x2+2y2=1
Let coordinate of D (22cosθ+4,22sinθ+3)≡(h,k)
So, 22h−4=cosθ...(1)
And 22k−3=sinθ...(2)
Equation (1)2+equation (2)2, then we get
(22h−4)2+(22k−3)2=1⇒1(x−2)2+(21)(y−23)2=1
∴ Required eccentricity is e=1−21=21
The locus of the mid-point of the line segment joining the point (4,3) and the points on the ellipse x2+2y2=4 is an ellipse with eccentricity
Held on 26 Jun 2022 · Verified 6 Jul 2026.
23
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21
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