A) \[\pm \frac{\pi }{2}\]
B) \[\pm \pi \]
C) 0
D) None of thesew
Correct Answer: A
Solution :
Let \[y={{m}_{1}}x\]and \[y={{m}_{2}}x\] be a pair of conjugate diameters of an ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\] and let \[P(a\cos \theta ,\,b\sin \theta )\] and \[Q(a\cos \varphi ,\,b\sin \varphi )\] be ends of these two diameters. Then \[{{m}_{1}}{{m}_{2}}=-\frac{{{b}^{2}}}{{{a}^{2}}}\] \[\Rightarrow \frac{b\sin \theta -0}{a\cos \theta -0}\times \frac{b\sin \varphi -0}{a\cos \varphi -0}=-\frac{{{b}^{2}}}{{{a}^{2}}}\] Þ\[\sin \theta \sin \varphi =-\cos \theta \cos \varphi \]Þ\[\cos (\theta -\varphi )=0\Rightarrow \theta -\varphi =\pm \frac{\pi }{2}\].You need to login to perform this action.
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