Question

If $p$ is the length of the perpendicular from a focus upon the tangent at any point $P$ of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $r$ is the distance of $P$ from the focus, then $\frac{2a}{r}-\frac{b^2}{p^2}$ is equal to

• A. $1$
• B. $2$
• C. $3$
• D. $0$

Answer 1

The correct option is A $1$

Let the point P be $(a\cos \theta ,b\sin \theta )$

The equation of the tangent at P is $\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

Length of the perpendicular from focus $(ae,0)$ is $p$.

$p=\frac{ecos\theta -1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}=\frac{ab(ecos\theta -1)}{\sqrt{b^{2}co{s}^2\theta +a^2(1-co{s}^2\theta )}}=\frac{ab(ecos\theta -1)}{\sqrt{a^2-a^2{e}^{2}co{s}^2\theta }}=-b\sqrt{\frac{1-ecos\theta }{1+ecos\theta }}$

$\Rightarrow \frac{b^2}{p^2}=\frac{1+ecos\theta }{1-ecos\theta }$

Now $r^2=(ae-acos\theta {)}^2+b^{2}si{n}^2\theta =a^2[(e-cos\theta {)}^2+(1-e^2\left)si{n}^2\theta \right]=a^2[e^{2}co{s}^2\theta -2ecos\theta +1]=a^2(1-ecos\theta {)}^2$

$\Rightarrow r=a(1-ecos\theta )$

Now,

$\frac{2a}{r}-\frac{b^2}{p^2}=\frac{2}{1-ecos\theta }-\frac{1+ecos\theta }{1-ecos\theta }=\frac{1-ecos\theta }{1-ecos\theta }=1$

Therefore $\frac{2a}{r}-\frac{b^2}{p^2}=1$