Find the condition for the line ax + by + c = | vedclass

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(C) Let a point $P(2 \cos 	heta, 6 \sin 	heta)$ be on the ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$. The equation of the normal to the ellipse at

Vedclass · Accepted Answer

$\frac{1}{a^2} + \frac{9}{b^2} = \frac{256}{c^2}$

Vedclass · Answer

(C) Let a point $P(2 \cos 	heta, 6 \sin 	heta)$ be on the ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$. The equation of the normal to the ellipse at point $P$ is given by $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$. Here $a^2 = 4$ and $b^2 = 36$. Substituting the coordinates $(2 \cos 	heta, 6 \sin 	heta)$,we get: $\frac{4x}{2 \cos 	heta} - \frac{36y}{6 \sin 	heta} = 4 - 36$. Simplifying,we have $2x \sec 	heta - 6y \operatorname{cosec} 	heta = -32$,or $2x \sec 	heta - 6y \operatorname{cosec} 	heta + 32 = 0$. Comparing this with $ax + by + c = 0$,we have $\frac{a}{2 \sec 	heta} = \frac{b}{-6 \operatorname{cosec} 	heta} = \frac{c}{32}$. This gives $\cos 	heta = \frac{2c}{32a} = \frac{c}{16a}$ and $\sin 	heta = -\frac{6c}{32b} = -\frac{3c}{16b}$. Using $\cos^2 	heta + \sin^2 	heta = 1$,we get $\frac{c^2}{256a^2} + \frac{9c^2}{256b^2} = 1$,which simplifies to $\frac{1}{a^2} + \frac{9}{b^2} = \frac{256}{c^2}$.

Find the condition for the line $ax + by + c = 0$ to be a normal to an ellipse $\frac{x^2}{4} + \frac{y^2}{36} = 1$.

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