Question

If $ p,\,\,p' $ 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
$\frac{x}{a} \cos \theta+\frac{y}{b} \sin \theta=1 \ldots$ (i)
$\therefore p=$ perpendicular distance from focus $(a e, 0)$ to the line (i)
$=\frac{\left|\frac{\mid \epsilon}{a} \cos \theta+0-1\right|}{\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta^{-}}{b^{2}}}}$
$=\frac{1-e \cos \theta}{\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin 2 \theta}{b^{2}}}} . .$ (i)
Also, $p'=$ 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, $r=S P=a(1-e \cos \theta)$
$\therefore a p=\frac{a-a e \cos \theta}{\sqrt{\frac{\cos ^{2} \theta}{a^{2}}+\frac{\sin ^{2} \theta}{b^{2}}}}=r p'$