The product of the perpendiculars drawn from the two foci of an ellipse to the tangent at any point of the ellipse is

Options

A. `a^(2)`
B. `b^(2)`
C. `4a^(2)`
D. `4b^(2)`

Answer

B

Explanation

Let P `( a cos theta , b sin theta )` be any point on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` the equation of the perpendiculars form S (ae,0) and S' `(-ae,0) ` on (i) is given by
`P=|(aeb cos theta-ab)/(sqrt(b^(2))cos^(2) theta +a^(2)sin ^(2)theta)||(ae b cos theta + ab)/(sqrt(b^(2)) cos ^(2) theta +a^(2) sin ^(2) theta )|`
`implies P=(a^(2) b^(2) -a^(2) b^(2) e^(2) cos ^(2) theta)/( a^(2) sin ^(2) theta+ b^(2) cos^(2) theta )= (a^(2) b^(2) (1-e^(2) cos^(2) theta))/(a^(2) sin^(2) theta +a^(2) (1-e^(2))cos^(2) theta)`
`=P=(a^(2) b^(2)(1-e^(2) cos ^(2) theta))/(a^(2)(1-e^(2) cos ^(2) theta))=b^(2)`