The normal at a variable point P on an ellips | vedclass

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Question

Normal at point $P(a \cos 	heta, b \sin 	heta)$ is

$\frac{a x}{\cos 	heta}-\frac{b y}{\sin 	heta}=a^{2}-b^{2}$

It meets axes at $Q\left(\fra

Vedclass · Accepted Answer

$e' = e$

Vedclass · Answer

Normal at point $P(a \cos 	heta, b \sin 	heta)$ is

$\frac{a x}{\cos 	heta}-\frac{b y}{\sin 	heta}=a^{2}-b^{2}$

It meets axes at $Q\left(\frac{\left(a^{2}-b^{2}ight) \cos 	heta}{a}, 0ight)$

and $R\left(0,-\frac{\left(a^{2}-b^{2}ight) \sin 	heta}{b}ight)$

Let $T(h, k)$ is a midpoint of $Q R$ Then $2 h=\frac{\left(a^{2}-b^{2}ight) \cos 	heta}{a}$

and $2 k=-\frac{\left(a^{2}-b^{2}ight) \sin 	heta}{b}$

$\Rightarrow \cos ^{2} 	heta+\sin ^{2} 	heta=\frac{4 h^{2} a^{2}}{\left(a^{2}-b^{2}ight)^{2}}+\frac{4 k^{2} b^{2}}{\left(a^{2}-b^{2}ight)^{2}}=1$

$\Rightarrow$ Locus is $\frac{x^{2}}{\frac{\left(a^{2}-b^{2}ight)^{2}}{4 a^{2}}}+\frac{y^{2}}{\frac{\left(a^{2}-b^{2}ight)^{2}}{4 b^{2}}}=1$

which is an ellipse, having eccentricity $e^{\prime},$ given by $e^{r 2}=1-\frac{\frac{\left(a^{2}-b^{2}ight)^{2}}{4 a^{2}}}{\frac{\left(a^{2}-b^{2}ight)^{2}}{4 b^{2}}}=1-\frac{b^{2}}{a^{2}}=e^{2}$

$e^{\prime}=e$

Note : In Equation (ii), $\frac{\left(a^{2}-b^{2}ight)}{4 a^{2}}<\frac{\left(a^{2}-b^{2}ight)}{4 b^{2}}$. Hence, $x$ -axis is minor axis.

The normal at a variable point $P$ on an ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}= 1$ of eccentricity e meets the axes of the ellipse in $ Q$ and $R$ then the locus of the mid-point of $QR$ is a conic with an eccentricity $e' $ such that :

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