Ellipse Ques 29
- 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 ellipse, then show that
$ \left(P F _1-P F _2\right)^{2}=4 a^{2} \quad 1-\frac{b^{2}}{d^{2}} $
$(1995,5 M)$
Show Answer
Solution:
- 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=1 $……(i)
We have, $d=$ length of perpendicular from $O$ to the
$ \begin{aligned} d & =\frac{|0+0-1|}{\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}} \\ \Rightarrow \quad \frac{1}{d} & =\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}} \\ \Rightarrow \quad \frac{1}{d^{2}} & =\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}} \end{aligned} $
We have to prove $\left(P F _1-P F _2\right)^{2}=4 a^{2} \quad (1-\frac{b^{2}}{d^{2}})$
Now, $\quad$ RHS $=4 a^{2} \quad (1-\frac{b^{2}}{d^{2}})=4 a^{2}-\frac{4 a^{2} b^{2}}{d^{2}}$
$ \begin{aligned} &=4 a^{2}-4 a^{2} b^{2} (\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}) \\ &=4 a^{2}-4 b^{2} \cos ^{2} \theta-4 a^{2} \sin ^{2} \theta \\ &=4 a^{2}\left(1-\sin ^{2} \theta\right)-4 b^{2} \cos ^{2} \theta \\ &=4 a^{2} \cos ^{2} \theta-4 b^{2} \cos ^{2} \theta \\ &=4 \cos ^{2} \theta\left(a^{2}-b^{2}\right)=4 \cos ^{2} \theta \cdot a^{2} e^{2} \quad \because e=\sqrt{1-(b / a)^{2}} \end{aligned} $
Again, $P F _1=e|a \cos \theta+a / e|=a|e \cos \theta+1|$
$ =a(e \cos \theta+1) $
$[\because-1 \leq \cos \theta \leq 1$ and $0<e<1]$
Similarly, $\quad P F _2=a(1-e \cos \theta)$
Therefore, $LHS=\left(P F _1-P F _2\right)^{2}$
$ \begin{aligned} & =[a(e \cos \theta+1)-a(1-e \cos \theta)]^{2} \\ & =(a e \cos \theta+a-a+a e \cos \theta)^{2} \\ & =(2 a e \cos \theta)^{2}=4 a^{2} e^{2} \cos ^{2} \theta \end{aligned} $
Hence, $\quad$ LHS $=$ RHS
BETA
