The locus of the poles of normal chords of an | vedclass

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(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(i)$.Let $(h, k)$ be the pole of the chord.The equation of the polar o

Vedclass · Accepted Answer

$\frac{a^6}{x^2} + \frac{b^6}{y^2} = (a^2 - b^2)^2$

Vedclass · Answer

(A) Let the equation of the ellipse be $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(i)$.Let $(h, k)$ be the pole of the chord.The equation of the polar of $(h, k)$ with respect to the ellipse is $\frac{xh}{a^2} + \frac{yk}{b^2} = 1$ $(ii)$.If this chord is a normal at point $	heta$,its equation is $ax \sec 	heta - by \csc 	heta = a^2 - b^2$ $(iii)$.Comparing $(ii)$ and $(iii)$,we have:$\frac{h/a^2}{a \sec 	heta} = \frac{k/b^2}{-b \csc 	heta} = \frac{1}{a^2 - b^2}$.This gives $\cos 	heta = \frac{a^3}{h(a^2 - b^2)}$ and $\sin 	heta = \frac{-b^3}{k(a^2 - b^2)}$.Using $\cos^2 	heta + \sin^2 	heta = 1$,we get:$\frac{a^6}{h^2(a^2 - b^2)^2} + \frac{b^6}{k^2(a^2 - b^2)^2} = 1$.Thus,the locus of $(h, k)$ is $\frac{a^6}{x^2} + \frac{b^6}{y^2} = (a^2 - b^2)^2$.

The locus of the poles of normal chords of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by:

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