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

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Question

(c) Let $P$ $(a\cos 	heta ,\,b\sin 	heta )$ be a point on the ellipse

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.$ Then the equation

Vedclass · Accepted Answer

${b^2}:{a^2}$

Vedclass · Answer

(c) Let $P$ $(a\cos 	heta ,\,b\sin 	heta )$ be a point on the ellipse

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.$ Then the equation of the normal at $P$ is

$ax\sec 	heta - by\,\,{m{cosec}}	heta = {a^2} - {b^2}$.

It meets the co-ordinate axes at $G\,\left( {\frac{{{a^2} - {b^2}}}{a}\cos 	heta ,\,0} ight)$ and

$g\,{m{ }}\left( {0, - \frac{{{a^2} - {b^2}}}{b}\sin 	heta } ight)$.

$ \Rightarrow P{G^2} = {\left( {a\cos 	heta - \frac{{{a^2} - {b^2}}}{a}\cos 	heta } ight)^2} + {b^2}{\sin ^2}	heta $

$ = \frac{{{b^2}}}{{{a^2}}}({b^2}{\cos ^2}	heta + {a^2}{\sin ^2}	heta )$

and $P{g^2} = \frac{{{a^2}}}{{{b^2}}}({b^2}{\cos ^2}	heta + {a^2}{\sin ^2}	heta )$, $	herefore$ $PG:Pg = {b^2}:{a^2}$.

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

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