The equation of the normal at the point (a se | vedclass

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(C) The given curve is $b^2 x^2 - a^2 y^2 = a^2 b^2$,which can be written as $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Differentiating with respect to $

Vedclass · Accepted Answer

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

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(C) The given curve is $b^2 x^2 - a^2 y^2 = a^2 b^2$,which can be written as $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Differentiating with respect to $x$,we get $\frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = \frac{b^2 x}{a^2 y}$.At the point $(a \sec 	heta, b 	an 	heta)$,the slope of the tangent is $m_t = \frac{b^2 (a \sec 	heta)}{a^2 (b 	an 	heta)} = \frac{b \sec 	heta}{a 	an 	heta}$.The slope of the normal is $m_n = -\frac{1}{m_t} = -\frac{a 	an 	heta}{b \sec 	heta}$.The equation of the normal is $y - b 	an 	heta = -\frac{a 	an 	heta}{b \sec 	heta} (x - a \sec 	heta)$.Multiplying by $b \sec 	heta$,we get $by \sec 	heta - b^2 \sec 	heta 	an 	heta = -ax 	an 	heta + a^2 \sec 	heta 	an 	heta$.Rearranging terms,$ax 	an 	heta + by \sec 	heta = (a^2 + b^2) \sec 	heta 	an 	heta$.Dividing by $\sec 	heta 	an 	heta$,we get $\frac{ax}{\sec 	heta} + \frac{by}{	an 	heta} = a^2 + b^2$.

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

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