If the normal at any point P on the ellipse f | vedclass

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(C) Let $P(a \cos 	heta, b \sin 	heta)$ be a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.The equation of the normal at $P$ is given

Vedclass · Accepted Answer

$b^2:a^2$

Vedclass · Answer

(C) Let $P(a \cos 	heta, b \sin 	heta)$ be a point on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.The equation of the normal at $P$ is given by $ax \sec 	heta - by \csc 	heta = a^2 - b^2$.This normal meets the $x$-axis at $G\left(\frac{a^2 - b^2}{a} \cos 	heta, 0ight)$ and the $y$-axis at $g\left(0, -\frac{a^2 - b^2}{b} \sin 	hetaight)$.The distance $PG$ is calculated as $PG = \sqrt{(a \cos 	heta - \frac{a^2 - b^2}{a} \cos 	heta)^2 + (b \sin 	heta - 0)^2} = \sqrt{(\frac{b^2}{a} \cos 	heta)^2 + b^2 \sin^2 	heta} = \frac{b}{a} \sqrt{b^2 \cos^2 	heta + a^2 \sin^2 	heta}$.The distance $Pg$ is calculated as $Pg = \sqrt{(a \cos 	heta - 0)^2 + (b \sin 	heta + \frac{a^2 - b^2}{b} \sin 	heta)^2} = \sqrt{a^2 \cos^2 	heta + (\frac{a^2}{b} \sin 	heta)^2} = \frac{a}{b} \sqrt{b^2 \cos^2 	heta + a^2 \sin^2 	heta}$.Therefore,the ratio $PG:Pg = \frac{b/a}{a/b} = \frac{b^2}{a^2}$.

If the normal at any point $P$ on the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ meets the coordinate axes in $G$ and $g$ respectively,then $PG:Pg = $

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