The equation of the normal at the point \[(a\sec \theta ,\ b\tan \theta )\] of the curve \[{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\] is [Karnataka CET 1999]

\[\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }={{a}^{2}}-{{b}^{2}}\]

\[\frac{ax}{\cos \theta }+\frac{by}{\sin \theta }={{a}^{2}}+{{b}^{2}}\]

\[\frac{ax}{\tan \theta }+\frac{by}{\sec \theta }={{a}^{2}}+{{b}^{2}}\]

\[\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }={{a}^{2}}+{{b}^{2}}\]

The equation of the normal at the point \[(a\sec \theta ,\ b\tan \thet

1.	The equation of the normal at the point \[(a\sec \theta ,\ b\tan \theta )\] of the curve \[{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\] is [Karnataka CET 1999]
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.	\[\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }={{a}^{2}}-{{b}^{2}}\]
Answer» D. \[\frac{ax}{\sec \theta }+\frac{by}{\tan \theta }={{a}^{2}}-{{b}^{2}}\]

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The equation of the normal at the point \[(a\sec \theta ,\ b\tan \theta )\] of the curve \[{{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}\] is [Karnataka CET 1999]

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