Question

If $p,p^{\prime }$ denote the lengths of the perpendiculars from the focus and the centre of an ellipse whose semi major axis is of length $a$ units on a tangent at a point on the ellipse and $r$ denotes the focal distance of the point, then

• A. $rp=a{p}^{\prime }$
• B. $rp+1=a{p}^{\prime }$
• C. $ap=r{p}^{\prime }$
• D. $ap=r{p}^{\prime }-1$

Answer 1

The correct option is C $ap=r{p}^{\prime }$

Tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $P(a\cos \theta ,b\sin \theta )$ is
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1$

$\therefore p=$ distance of a focus$(ae,0)$ from the tangent
$\Rightarrow p=\frac{|\frac{ae}{a}\cos \theta +0-1|}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}$
$\Rightarrow p=\frac{1-e\cos \theta }{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}$

And $p^{\prime }=$distance of centre$(0,0)$ from the tangent
$\Rightarrow p^{\prime }=\frac{1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}$

Also $r=SP=a(1\pm e\cos \theta )$
where $S$ is any focus of ellipse.

$\therefore ap=\frac{a-ae\cos \theta }{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}=r{p}^{\prime }$