Prove that the chords of contact of pairs of perpendicular tangents to the ellipse `x^2/a^2+y^2/b^2=1` touch another fixed ellipse.

Options

A. `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =(1)/((2a^(2)+b^(2)))`
B. `(x^(2))/(a^(2))+(y^(2))/(b^(2)) =(2)/((a^(2)-b^(2)))`
C. `(x^(2))/(a^(4))+(y^(2))/(b^(4)) =(1)/((a^(2)+b^(2)))`
D. `(x^(2))/(a^(2))-(y^(2))/(b^(2)) =(2)/((3a^(2)-b^(2)))`

Answer

C

Explanation

We known that locus of the point of intersection of perpendicular tangents to the given ellipse is `x^(2) + y^(2) = a^(2) + b^(2)`. Any point on this circle can be taken as
`P -= (sqrt(a^(2)+b^(2)) cos theta, sqrt(a^(2)+b^(2)) sin theta)`
The equation of the chord of contact of tangents from P is
`(x)/(a^(2)) sqrt(a^(2)+b^(2)) cos theta +(y)/(b^(2)) sqrt(a^(2)+b^(2)) sin theta =1`.
Let this line be a tangent to the fixed ellipse `(x^(2))/(A^(2)) +(y^(2))/(B^(2)) =1`.
`rArr (x)/(A) cos theta +(y)/(B) sin theta =1`,
Where `A = (a^(2))/(sqrt(a^(2)+b^(2))), B = (b^(2))/(sqrt(a^(2)+b^(2)))`
`(x^(2))/(a^(4)) + (y^(2))/(b^(4)) = (1)/((a^(2)+b^(2)))`