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

Options

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

Answer

B

Explanation

Let (h,k) be the circumference of `DeltaPQR`.
Perpendicular tangents are drawn from a point P to the ellipse `x^(2)+4y^(2)=4`. So, P lies on the director circle `x^(2)+y^(2)=5`. Let the coordinates of P be `(sqrt5 cos theta, sqrt5 sin theta)`. Then, the equation of the chord of contact of tangents drawn from P to the ellipse is
`sqrt5x costheta+4sqrt5y sintheta=4" "...(i)`
Since(h,k) are the coordinates of the mid-point of QR.
So, equation of QR is
`hx+4ky=h^(2)+4k^(2)" "[" Using "S=T]" "...(ii)`
Clearly, (i) and (ii) represent the same line.
`therefore(sqrt5 cos theta)/(h)=(4sqrt5sintheta)/(4k)=(4)/(h^(2)+4k^(2))`
`rArr costheta = (4h)/(sqrt5(h^(2)+4k^(2))),sintheta=(4k)/(sqrt5(h^(2)+4k^(2)))`
`rArr(16)/(5)(h^(2)+k^(2))=(h^(2)+4k^(2))^(2)`
Hence, the locus of (h ,k) is `x^(2)+y^(2)=(5)/(16)=(x^(2)+4y^(2))^(2)`