A) \[\frac{{{a}^{2}}+{{b}^{2}}}{a}\]
B) \[-\left( \frac{{{a}^{2}}+{{b}^{2}}}{a} \right)\]
C) \[\frac{{{a}^{2}}+{{b}^{2}}}{b}\]
D) \[-\left( \frac{{{a}^{2}}+{{b}^{2}}}{b} \right)\]
Correct Answer: D
Solution :
[d] Equation of the normal to the hyperbola\[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] at the point \[(aseca,btana)\] is given by \[ax\cos \alpha +by\cot \alpha ={{a}^{2}}+{{b}^{2}}\] Normal at \[\theta ,\phi \] are \[\left\{ \begin{matrix} ax\cos \theta +by\cot \theta ={{a}^{2}}+{{b}^{2}} \\ ax\cos \phi +by\cot \phi ={{a}^{2}}+{{b}^{2}} \\ \end{matrix} \right.\] where \[\phi =\frac{\pi }{2}-\theta \] and these passes through \[(h,k)\] \[\therefore \,\,\,\,ah\cos \theta +bk\cot \theta ={{a}^{2}}+{{b}^{2}}\] \[\Rightarrow ah\sin \theta +bk\tan \theta ={{a}^{2}}+{{b}^{2}}\] Eliminating \[h,\,\,bk(cot\theta sin\theta -tan\theta cos\theta )\] \[=({{a}^{2}}+{{b}^{2}})(sin\theta -cos\theta )\] or \[k=-({{a}^{2}}+{{b}^{2}})/b\]You need to login to perform this action.
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