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 $\left( PF _{1}- PF _{2}\right)^{2}=$

• A. $4 a^{2}\left(1-\frac{b^{2}}{d^{2}}\right)$
• B. $a^{2}\left(1-\frac{b^{2}}{d^{2}}\right)$
• C. $4 b^{2}\left(1-\frac{a^{2}}{d^{2}}\right)$
• 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,
$\left( PF _{1}- PF _{2}\right)^{2}=(2 a e \cos \theta)^{2}=4 a ^{2} e ^{2} \cos ^{2} \theta=4 a ^{2}\left(1-\frac{ b ^{2}}{ d ^{2}}\right)$