A) \[\frac{4}{{{x}^{2}}}+\frac{2}{{{y}^{2}}}=1\]
B) \[\frac{2}{{{x}^{2}}}-\frac{4}{{{y}^{2}}}=1\]
C) \[\frac{2}{{{x}^{2}}}+\frac{4}{{{y}^{2}}}=1\]
D) \[\frac{4}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=1\]
Correct Answer: D
Solution :
Equation of the tangent at the point '\[\theta \]' is \[\frac{x\sec \theta }{a}-\frac{y\tan \theta }{b}=1\] \[\Rightarrow \]\[P=(a\cos \theta ,0)\]and\[Q=(0,-b\cot \theta )\] Let R be\[(h,k)\Rightarrow h=a\cos \theta ,k=-b\cot \theta \] \[\Rightarrow \]\[\frac{k}{h}=\frac{-b}{a\sin \theta }\Rightarrow \sin \theta =\frac{-bh}{ak}\]and\[\cos \theta =\frac{h}{a}\] By squaring and adding,\[\frac{{{b}^{2}}{{h}^{2}}}{{{a}^{2}}{{k}^{2}}}+\frac{{{h}^{2}}}{{{a}^{2}}}=1\]You need to login to perform this action.
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