• # question_answer The locus of the poles of normal chords of an ellipse is given by  [UPSEAT 2001] A)            $\frac{{{a}^{6}}}{{{x}^{2}}}+\frac{{{b}^{6}}}{{{y}^{2}}}={{({{a}^{2}}-{{b}^{2}})}^{2}}$             B)            $\frac{{{a}^{3}}}{{{x}^{2}}}+\frac{{{b}^{3}}}{{{y}^{2}}}={{({{a}^{2}}-{{b}^{2}})}^{2}}$ C)            $\frac{{{a}^{6}}}{{{x}^{2}}}+\frac{{{b}^{6}}}{{{y}^{2}}}={{({{a}^{2}}+{{b}^{2}})}^{2}}$            D)            $\frac{{{a}^{3}}}{{{x}^{2}}}+\frac{{{b}^{3}}}{{{y}^{2}}}={{({{a}^{2}}+{{b}^{2}})}^{2}}$

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 $ax\,\sec \theta -\,by\,\text{cosec}\,\theta ={{a}^{\text{2}}}-{{b}^{2}}$                       .....(iii)            Hence comparing (ii) and (iii), we get            $\frac{(h/{{a}^{2}})}{a\,\sec \theta }=\frac{(k/{{b}^{2}})}{-b\,\cos ec\theta }=\frac{1}{({{a}^{2}}-{{b}^{2}})}$            Þ $\cos \theta =\frac{{{a}^{3}}}{h\,({{a}^{2}}-{{b}^{2}})}$ and $\sin \theta =\frac{{{b}^{3}}}{k({{a}^{2}}-{{b}^{2}})}$            Squaring and adding we get, $1=\frac{1}{{{({{a}^{2}}-{{b}^{2}})}^{2}}}\left( \frac{{{a}^{6}}}{{{h}^{2}}}+\frac{{{b}^{6}}}{{{k}^{2}}} \right)\,$                    \ Required locus of $(h,\,k)$ is $\frac{{{a}^{6}}}{{{x}^{2}}}+\frac{{{b}^{6}}}{{{y}^{2}}}={{({{a}^{2}}-{{b}^{2}})}^{2}}.$.