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The ratio of the area enclosed by the lo...

The ratio of the area enclosed by the locus of the midpoint of PS and area of the ellipse is (P-be any point on the ellipse and S, its focus)

A

A. `(1)/(2)`

B

B. `(1)/(3)`

C

C. `(1)/(5)`

D

D. `(1)/(4)`

Text Solution

Verified by Experts

The correct Answer is:
D
Let ellipse be `(x^(2))/(a^(2)) +(y^(2))/(b^(2)) =1` Mid point of PS is (h,k)
`:. h = (a cos theta +ae)/(2) rArr cos theta = (2h-ae)/(a)`
and `k = (b sin theta)/(2)`
Eliminating `theta`, we get locus as
`((2h-ae)^(2))/(a^(2)) +(4k^(2))/(b^(2)) =1`
`rArr ((x-(ae)/(2)))/((a^(2))/(4)) +(y^(2))/((b^(2))/(4))=1`, which is ellipse having enclosed area
Area `rArr pi(a)/(2).(b)/(2) = (piab)/(4). :.` ratio `= (1)/(4)`
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