Conic Sections Question 505
Question: The locus of the poles of normal chords of an ellipse is given by
[UPSEAT 2001]
Options:
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}} $
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Answer:
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 -bycosec\theta ={a^{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})} $
therefore $ \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}}}( \frac{a^{6}}{h^{2}}+\frac{b^{6}}{k^{2}} ) $
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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