The equation of the normal to the curve x^{2} | vedclass

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(A) The equation of the circle is $x^{2} + y^{2} = r^{2}$.Any point $P$ on the circle can be represented as $(r \cos 	heta, r \sin 	heta)$.The slope

Vedclass · Accepted Answer

$x \sin 	heta - y \cos 	heta = 0$

Vedclass · Answer

(A) The equation of the circle is $x^{2} + y^{2} = r^{2}$.Any point $P$ on the circle can be represented as $(r \cos 	heta, r \sin 	heta)$.The slope of the tangent at $P$ is given by differentiating $x^{2} + y^{2} = r^{2}$ with respect to $x$:$2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$.At point $P(r \cos 	heta, r \sin 	heta)$,the slope of the tangent $m_{t} = -\frac{r \cos 	heta}{r \sin 	heta} = -\cot 	heta$.The slope of the normal $m_{n}$ is $-\frac{1}{m_{t}} = \frac{1}{\cot 	heta} = 	an 	heta = \frac{\sin 	heta}{\cos 	heta}$.The equation of the normal at $(r \cos 	heta, r \sin 	heta)$ is:$y - r \sin 	heta = \frac{\sin 	heta}{\cos 	heta} (x - r \cos 	heta)$.$y \cos 	heta - r \sin 	heta \cos 	heta = x \sin 	heta - r \sin 	heta \cos 	heta$.$x \sin 	heta - y \cos 	heta = 0$.

The equation of the normal to the curve $x^{2}+y^{2}=r^{2}$ at the point $P(r \cos \theta, r \sin \theta)$ is:

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