Question

If the normal at any point $P$ on an ellipse of eccentricity ${}^{\prime }{e}^{\prime }$ meets the major axis and minor axis at $T$ and $G$ respectively, then $PT:PG$ equal to

• A. $e$
• B. $1:e$
• C. $1-e^2:1$
• D. $1-e^2:e$

Answer 1

Answer: (C)

$1-e^2:1$

Ellipse $\to \frac{x^2}{a^2}+\frac{y^2}{6^2}=1$

Equation of normal at point $P(a\cos \theta ,b\sin \theta )$

is $ax\sec \theta -6y\csc \theta =a^2-b^2$

$9+$ cuts on $x-$axis

$x=\frac{a^2-b^2}{a\sec \theta }T(\frac{a^2-b^2}{a\sec \theta },0)$

$9+$ cuts $y-$axis at

$y=a\frac{a^2-b^2}{-b\sin \theta }Cr.(0,\frac{b^2-a^2}{b\csc \theta })$

$PT=\sqrt{(a\cos \theta -{\left(\frac{a^2-b^2}{a}\right)}^2+6^2{\sin }^2\theta )}$

$PG=\sqrt{a^2{\cos }^2\theta +{\left(b\sin \theta \left(\frac{b^2-a^2}{b\csc \theta }\right)\right)}^2}$

$\frac{PT}{PG}=\sqrt{\frac{b^4}{a^2}{\cos }^2\theta +6^2{\sin }^2\theta }$

$\sqrt{a^2{\cos }^2\theta +\frac{a^4}{b^2}{\sin }^2\theta }$

$\frac{PT}{PG}=\sqrt{\frac{(1-e^2)b^2{\cos }^2\theta +b^2{\sin }^2\theta }{\left(\frac{1-a^2}{1-e^2}\right){\sin }^2\theta +a^2{\cos }^2\theta }}=\sqrt{1-e^2}\sqrt{\frac{b^2-e^2{b}^2{\cos }^2\theta }{b^2-e^2{e}^2{\cos }^2\theta }}$

$\frac{PT}{PG}=\sqrt{1-e^2}\sqrt{\frac{b^2}{b^2}}=\sqrt{1-e^2}\sqrt{1-e^2}=1-e^2:1$