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 $F_1$ and $F_2$ are the two foci of the ellipsë, then show that
${\left(P{F}_1-P{F}_2\right)}^2=4a^2\left(1-\frac{b^2}{d^2}\right)$

Answer 1

Let the coordinates of point $P$ be $(a\cos \theta ,b\sin \theta )$.
Then, equation of tangent at $P$ is
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =\mathrm{1....}$(i)
We have, $d=$ length of perpendicular from $O$ to the tangent at $P$

$d=\frac{|0+0-1|}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}\Rightarrow \frac{1}{d}=\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}$
$\Rightarrow \frac{1}{d^2}=\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}$
We have to prove ${\left(P{F}_1-P{F}_2\right)}^2=4a^2\left(1-\frac{b^2}{d^2}\right)$
Now, $RHS=4a^2\left(1-\frac{b^2}{d^2}\right)$
$=4a^2-\frac{4a^2{b}^2}{d^2}$
$=4a^2-4a^2{b}^2\left(\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}\right)$
$=4a^2-4b^2{\cos }^2\theta -4a^2{\sin }^2\theta$
$=4a^2\left(1-{\sin }^2\theta \right)-4b^2{\cos }^2\theta$
$=4a^2{\cos }^2\theta -4b^2{\cos }^2\theta$
$=4{\cos }^2\theta \left(a^2-b^2\right)=4{\cos }^2\theta \cdot a^2{e}^2\left[\because e=\sqrt{1-(b/a{)}^2}\right]$
Again, $P{F}_1=e|a\cos \theta +a|e\mid$
$=a|e\cos \theta +1|$
$=a(e\cos \theta +1)$
$[\because -1\le \cos \theta \le 1$ and $0<e<1]$
Similarly, $P{F}_2=a(1-e\cos \theta )$
Therefore, LHS $={\left(P{F}_1-P{F}_2\right)}^2$
$=[a(e\cos \theta +1)-a(1-e\cos \theta ){]}^2$
$=(ae\cos \theta +a-a+ae\cos \theta {)}^2$
$=(2ae\cos \theta {)}^2=4a^2{e}^2{\cos }^2\theta$
Hence, LHS $=$ RHS