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From a point P perpendicular tangents PQ and PR are drawn to ellipse `x^(2)+4y^(2) =4`, then locus of circumcentre of triangle PQR is

A

`x^(2)+y^(2) =(16)/(5)(x^(2)+4y^(2))^(2)`

B

`x^(2)+y^(2) =(5)/(16)(x^(2)+4y^(2))^(2)`

C

`x^(2)+4y^(2)=(16)/(5)(x^(2)+y^(2))^(2)`

D

`x^(2)+4y^(2)=(16)/(5)(x^(2)+y^(2))^(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
P lies on `x^(2) +y^(2) =5` (director circle of given ellipse)
Let `P(sqrt(5) cos theta, sqrt(5) sin theta)`
Since `DeltaQPR` is right angled at point P, circumcenter is mid-point of QR.
QR is chord of contact w.r.t point P,
`:.` Its equation is `(xsqrt(5)cos theta)/(4) +(y sqrt(5) sin theta)/(1) =1` (i)
Also equation of chord QR which is bisected at point (h,k) is
`(xh)/(h^(2) + 4k^(2)) +(4ky)/(h^(2)+4k^(2)) =1` (using `T = S_(1))` (ii)
Comparing coefficients of equations (i) and (ii),
`(sqrt(5)cos theta)/(4) = (h)/(h^(2) + 4k^(2))` and `sqrt(5) sin theta = (4k)/(h^(2)+4k^(2))`
`rArr x^(2)+y^(2) = (5)/(16) (x^(2) + 4y^(2))^(2)`
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