A) \[20/3\]
B) \[25/3\]
C) \[10\]
D) \[13/4\]
Correct Answer: B
Solution :
[b] The equation of normal at \[P(2\sec \theta ,\,\tan \theta )\]on hyperbola \[\frac{{{x}^{2}}}{4}-\frac{{{y}^{2}}}{1}=1\] is \[\frac{2x}{\sec \theta }+\frac{y}{\tan \theta }=5\] \[\therefore \] Slope of normal \[=\frac{-2\tan \theta }{\sec \theta }\] ...?(1) According to the question, \[\frac{-2\tan \theta }{\sec \theta }=-1\] \[\Rightarrow \,\,\,\sin \theta =\frac{1}{2}\] \[\Rightarrow \,\,\,\theta =\frac{\pi }{6}\] So, equation of normal becomes \[y=-x+\frac{5}{\sqrt{3}}\] This touches the ellipse \[\frac{{{x}^{2}}}{{{a}^{2}}}+\frac{{{y}^{2}}}{{{b}^{2}}}=1.\] \[\Rightarrow \,\,\,\,\frac{25}{3}={{a}^{2}}{{(-1)}^{2}}+{{b}^{2}}\] \[\Rightarrow \,\,\,\,{{a}^{2}}+{{b}^{2}}=\frac{25}{3}\]
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