A) \[\frac{4}{{{x}^{2}}}+\frac{2}{{{y}^{2}}}=1\]
B) \[\frac{4}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=1\]
C) \[\frac{2}{{{x}^{2}}}-\frac{4}{{{y}^{2}}}=1\]
D) \[\frac{2}{{{x}^{2}}}+\frac{4}{{{y}^{2}}}=1\]
Correct Answer: B
Solution :
\[\frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1\] \[\therefore P(a\cos \theta ,\,\,0)\,\,Q(0,\,\,-b\cot \theta )\] \[\therefore h=a\cos \theta \] \[\Rightarrow \sec \theta =\frac{a}{h}\] \[k=-b\cot \theta \] \[\Rightarrow \tan \theta =\frac{-b}{k}\] So, \[\frac{{{a}^{2}}}{{{x}^{2}}}-\frac{{{b}^{2}}}{{{y}^{2}}}=1\Rightarrow \frac{4}{{{x}^{2}}}-\frac{2}{{{y}^{2}}}=1\]You need to login to perform this action.
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