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

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Question

(a) Let the equation of the ellipse is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$.....$(i)$

Let $(h,\,k)$ be the poles.

Now 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 is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$.....$(i)$

Let $(h,\,k)$ be the poles.

Now polar of $(h,\,k)$ w.r.t. the ellipse is given by $\frac{{xh}}{{{a^2}}} + \frac{{yk}}{{{b^2}}} = 1$ .....$(ii)$

If it is a normal to the ellipse then it must be identical with $ax\,\sec 	heta - \,by\,{m{cosec}}\,	heta = {a^{m{2}}} - {b^2}$.....$(iii)$

Hence comparing $(ii)$ and $(iii),$ we get

$\frac{{(h/{a^2})}}{{a\,\sec 	heta }} = \frac{{(k/{b^2})}}{{ - b\,\cos ec	heta }} = \frac{1}{{({a^2} - {b^2})}}$

==> $\cos 	heta = \frac{{{a^3}}}{{h\,({a^2} - {b^2})}}$ and $\sin 	heta = \frac{{{b^3}}}{{k({a^2} - {b^2})}}$

Squaring and adding we get,

$1 = \frac{1}{{{{({a^2} - {b^2})}^2}}}\left( {\frac{{{a^6}}}{{{h^2}}} + \frac{{{b^6}}}{{{k^2}}}} ight)\,$

 Required 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 is given by

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