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

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Question

(C) The normal at point $P(a \cos 	heta, b \sin 	heta)$ is given by $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$.It meets the axes

Vedclass · Accepted Answer

$e' = e$

Vedclass · Answer

(C) The normal at point $P(a \cos 	heta, b \sin 	heta)$ is given by $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$.It meets the axes at $Q\left(\frac{(a^2 - b^2) \cos 	heta}{a}, 0ight)$ and $R\left(0, -\frac{(a^2 - b^2) \sin 	heta}{b}ight)$.Let $T(h, k)$ be the midpoint of $QR$. Then $2h = \frac{(a^2 - b^2) \cos 	heta}{a}$ and $2k = -\frac{(a^2 - b^2) \sin 	heta}{b}$.Using $\cos^2 	heta + \sin^2 	heta = 1$,we get $\frac{4h^2 a^2}{(a^2 - b^2)^2} + \frac{4k^2 b^2}{(a^2 - b^2)^2} = 1$.The locus is $\frac{x^2}{\frac{(a^2 - b^2)^2}{4a^2}} + \frac{y^2}{\frac{(a^2 - b^2)^2}{4b^2}} = 1$.This is an ellipse with eccentricity $e'$ given by $e'^2 = 1 - \frac{b^2}{a^2} = e^2$.Thus,$e' = e$.

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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