Question

Let d be the perpendicular distance from the centre of the ellipse `x^2/a^2+y^2/b^2=1` to the tangent drawn at a point P on the ellipse. If `F_1 & F_2` are the two foci of the ellipse, then show the `(PF_1-PF_2)^2=4a^2[1-b^2/d^2]`.

Answer 1

The equation of the tangent at the point `P(cos theta, b sin theta)` on the ellipse `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1` is
`(x)/(a) cos theta+(y)/(b) sin theta=1" "(1)`
The perpendicualr distance of (1) from the centre (0,0) of the ellipse is given by
`d=(1)/(sqrt((1)/(a^(2))cos^(2)theta+(1)/(b^(2))cos^(2)theta))=(ab)/(sqrt(b^(2)cos^(2)theta+a^(2)sin^(2)theta))`
`:. 4a^(2)(1-(b^(2))/(d^(2)))=4a^(2){1-(b^(2)cos^(2)theta+a^(2)sin^(2)theta)/(a^(2))}`
`4(a^(2)-b^(2))cos^(2)theta=4a^(2)e^(2)cos^(2)theta" "(2)`
The foci are `F_(1)-=(ae,0)and F_(2)-=(-ae),o)`. Therefore,
`PF_(1)=(1-ecostheta)`
and `PF_(2)=a(1+ e cos theta)`
`:. (PF_(1)-PF_(2))^(2)=4a^(2)e^(2)cos^(2)theta" "(3)`
Hence, from (2) and (3), we get have
`(PF_(1)-PF_(2))^(2)=4a^(2)(1-(b^(2))/(d^(2)))`