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 $F_{1}and{F}_2$ are the two foci of the ellipse then show that: $(P{F}_1-P{F}_2{)}^2=4a^2[1-\frac{b^2}{d^2}]$ .

Answer 1

Ellipse:$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1---\left(i\right)$

focus : $F_1:(ae,0)\&F_2:(-ae,0)$

centre:$(0,0)$

$M$ is perpendicular on tangent at $P(a\cos \varphi ,b\sin \varphi )$

equation of tangent $\Rightarrow \frac{x\cos \varphi }{a}+\frac{y\sin \varphi }{b}=1---\left(ii\right)$

length of $CM=|\frac{0+0-1}{\frac{\cos {}^2\varphi }{a^2}+\frac{\sin {}^2\varphi }{b^2}}|=\frac{a^2{b}^2}{b^2{\cos }^2\varphi +a^2{\sin }^2\varphi }=d---\left(iii\right)$

now, $P{F}_1=\sqrt{(a\cos \varphi -ae{)}^2+b^2\sin {\varphi }^2}=\sqrt{a^2{\cos }^2\varphi +b^2{\sin }^2\varphi +a^2{e}^2-2a^{2}e\cos \varphi }$

$P{F}_2=\sqrt{(a\cos \varphi +ae{)}^2+b^2\sin {\varphi }^2}=\sqrt{a^2{\cos }^2\varphi +b^2{\sin }^2\varphi +a^2{e}^2+2a^{2}e\cos \varphi }$

$\Rightarrow P{F}_1-P{F}_2=(0+..2\sqrt{(a^2{\cos }^2\varphi +b^2{\sin }^2\varphi +a^2{e}^2-2a^{2}e\cos \varphi )-(a^2{\cos }^2\varphi +b^2{\sin }^2\varphi +a^2{e}^2+2a^{2}e\cos \varphi )})$

$\Rightarrow 2.2a^2\sqrt{\frac{(b^2\cos {}^2\varphi +a^2{\sin }^2{\varphi )}^2-b^2(a^2{b}^2)}{\left(\right(b^2\cos {}^2\varphi +a^2{\sin }^2{\varphi )}^2)}}=+4a^2\sqrt{1-\frac{b^2}{d^2}}.......(d=\frac{a^2{b}^2}{b^2{\cos }^2\varphi +a^2{\sin }^2\varphi })$