A) \[\frac{ax}{\cos \theta }+\frac{by}{\sin \theta }={{a}^{2}}+{{b}^{2}}\]
B) \[\frac{ax}{\tan \theta }+\frac{by}{\sec \theta }={{a}^{2}}+{{b}^{2}}\]
C) \[\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }={{a}^{2}}+{{b}^{2}}\]
D) \[\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }={{a}^{2}}-{{b}^{2}}\]
Correct Answer: C
Solution :
Equation of normal to hyperbola \[\frac{{{x}^{2}}}{{{a}^{2}}}-\frac{{{y}^{2}}}{{{b}^{2}}}=1\] at \[(a\sec \theta ,b\tan \theta )\] is \[\frac{{{a}^{2}}x}{a\sec \theta }+\frac{{{b}^{2}}y}{b\tan \theta }={{a}^{2}}+{{b}^{2}}\].
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