A) 1
B) 2
C) 3
D) \[3\frac{1}{3}\]
Correct Answer: B
Solution :
Let\[P(a\cos \theta ,b\sin \theta )\]and\[Q(a\cos \phi ,b\sin \phi )\]be two points on the ellipse\[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1\]such that \[\theta -\phi =\frac{\pi }{2}\] Now, the equations of tangents and P and Q are respectively \[\frac{x}{a}\cos \theta +\frac{y}{b}\sin \theta =1\] ...(i) \[\frac{x}{a}\cos \phi +\frac{y}{b}\sin \phi =1\] ... (ii) Since,\[\theta =\frac{\pi }{2}+\phi \]so Eq. (i) reduces to \[-\frac{x}{a}\sin \phi +\frac{y}{b}\cos \phi =1\] ...(iii) Squaring and adding Eqs. (ii) and (iii), we get \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=2\] Hence, the points of intersection of tangents at P and Q lie on the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=2\]
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