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) \[\frac{{{a}^{3}}}{{{x}^{2}}}+\frac{{{b}^{3}}}{{{y}^{2}}}={{({{a}^{2}}+{{b}^{2}})}^{2}}\]
Correct Answer: A
Solution :
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 \[\frac{{{a}^{6}}}{{{x}^{2}}}+\frac{{{b}^{6}}}{{{y}^{2}}}={{({{a}^{2}}-{{b}^{2}})}^{2}}.\].
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