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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]`. |
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Answer» 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)))` |
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