The equation of the normal at the point $(a\sec heta ,\ b an heta )$ of the curve ${{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}$ is

Question

The equation of the normal at the point $(a\sec 	heta ,\ b	an 	heta )$ of the curve ${{b}^{2}}{{x}^{2}}-{{a}^{2}}{{y}^{2}}={{a}^{2}}{{b}^{2}}$ is

Accepted Answer

$\frac{ax}{\sec 	heta }+\frac{by}{	an 	heta }={{a}^{2}}+{{b}^{2}}$

Answer

$\frac{ax}{\cos 	heta }+\frac{by}{\sin 	heta }={{a}^{2}}+{{b}^{2}}$

Answer

$\frac{ax}{	an 	heta }+\frac{by}{\sec 	heta }={{a}^{2}}+{{b}^{2}}$

Answer

$\frac{ax}{\sec 	heta }+\frac{by}{	an 	heta }={{a}^{2}}-{{b}^{2}}$

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