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 :
[b] For hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\]equation of tangent at point \[P(a\,\sec \theta ,\,\,b\,\tan \theta )\] is \[\frac{x}{a}\sec \theta -\frac{y}{b}\tan \theta =1\] \[\therefore \,\,\,P(a\cos \theta ,0)\] and \[Q\,\,(0,-b\cot \,\theta )\] \[\therefore \,\,\,h=a\cos \theta \,\,\,\,\Rightarrow \,\,\,\sec \theta =\frac{a}{h}\] \[k=-b\cot \theta \,\,\,\,\Rightarrow \,\,\,\,\tan \theta =\frac{-b}{k}\] 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.
You will be redirected in
3 sec
