Question

Let $d$ be the perpendicular distance from the centre of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ to the tangent drawn at a point $P$ on the ellipse. If $F_1$ and $F_2$ be the foci of the ellipse, then ${\left(P{F}_1-P{F}_2\right)}^2=$

• A. $4a^2\left(1-\frac{b^2}{d^2}\right)$
• B. $a^2\left(1-\frac{b^2}{d^2}\right)$
• C. $4b^2\left(1-\frac{a^2}{d^2}\right)$
• D. $b^2\left(1-\frac{a^2}{d^2}\right)$

Answer 1

Answer: (A)

Let the point $P$ be $(a\cos \theta ,b\sin \theta )$ The equation of tangent at $P$ is
$\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1...(1)$
If $d$ be the length of perpendicular from the centre $C(0,0)$ of the ellipse to the tangent given by $(1)$ then
$d=\frac{1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\Rightarrow \frac{1}{d^2}=\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}$
$\Rightarrow \frac{b^2}{d^2}=\frac{b^2}{a^2}{\cos }^2\theta +1-{\cos }^2\theta$
$\Rightarrow 1-\frac{b^2}{d^2}=\left(1-\frac{b^2}{a^2}\right){\cos }^2\theta =e^2{\cos }^2\theta ...(2)$
Now,
${\left(P{F}_1-P{F}_2\right)}^2=(2ae\cos \theta {)}^2=4a^2{e}^2{\cos }^2\theta =4a^2\left(1-\frac{b^2}{d^2}\right)$