Question

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

• A. ap' = rp + 1
• B. rp = ap'
• C. ap = rp' + 1
• D. ap = rp'

Answer 1

Answer: (D)

Tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at $(a\cos \theta ,b\sin \theta )$ is
$\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\dots$ (i)
$\therefore p=$ perpendicular distance from focus $(ae,0)$ to the line (i)
$=\frac{|\frac{\mid ϵ}{a}\cos \theta +0-1|}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2{\theta }^{-}}{b^2}}}$
$=\frac{1-e\cos \theta }{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{\sin 2\theta }{b^2}}}..$ (i)
Also, $p^{\prime }=$ perpendicular distance from centre $(0,0)$ to the line (i)
$=\frac{1}{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}$ (ii)
Again, $r=SP=a(1-e\cos \theta )$
$\therefore ap=\frac{a-ae\cos \theta }{\sqrt{\frac{{\cos }^2\theta }{a^2}+\frac{{\sin }^2\theta }{b^2}}}=r{p}^{\prime }$