A line intesects the ellipse `(x^(2))/(4a^(2))+(y^(2))/(a^(2))=1` at A and B and the parabola `y^(2)=4a(x+2a)` at C and D. The line segment AB substends a right angle at the centre of the ellipse. Then, the locus of the point of intersection of tangents to the parabola at C and D, is

Options

A. `y^(2)-a^(2)=(5)/(4)(x-4a)^(2)`
B. `y^(2)-2a^(2)=10(x-4a)^(2)`
C. `y^(2)+a^(2)=(5)/(2)(x-4a)^(2)`
D. `y^(2)+4a^(2)=5(x+4a)^(2)`

Answer

D

Explanation

Let P(h,k)be the point of intersection of tangents to the parabola `y^(2)=4a(x+2a)` at C and D. Then , CD is the chord of contact of tangents to the parabola drawn from P and hence its equation is
`ky=2a(x+h)+8a^(2
)or,ky-2ax=2a(h+4a)`
The combined equations of the lines joining the origin (centre of the ellipse) to the points of intersection of line (i) and the ellipse `(x^(2))/(4a^(2))+(y^(2))/(a^(2))=1` is
`(x^(2))/(4a^(2))+(y^(2))/(a^(2))={(ky-2ax)/(2a(h+4a)^(2))}^(2)`
Lines given by the above equation are at right angle.
`therefore(1)/(4a^(2))-(1)/((h+4a)^(2))+(1)/(a^(2))-(k^(2))/(4a^(2)(h+4a)^(2))=0`
or, `k^(2)+4a^(2)=(5h+4a)^(2)`
Hence, the locus of (h,k) "is " `y^(2)+4a^(2)=5(x+4a)^(2)`.