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(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Differentiating with respect to $x$,we get $\frac{2x}{a^2} + \frac{2y}{b^2

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

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(C) The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Differentiating with respect to $x$,we get $\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0$,which implies $\frac{dy}{dx} = -\frac{b^2 x}{a^2 y}$. At the point $(a \cos 	heta, b \sin 	heta)$,the slope of the tangent is $m_t = -\frac{b^2 (a \cos 	heta)}{a^2 (b \sin 	heta)} = -\frac{b \cos 	heta}{a \sin 	heta}$. The slope of the normal is $m_n = -\frac{1}{m_t} = \frac{a \sin 	heta}{b \cos 	heta}$. The equation of the normal is $y - b \sin 	heta = \frac{a \sin 	heta}{b \cos 	heta} (x - a \cos 	heta)$. Multiplying by $b \cos 	heta$,we get $by \cos 	heta - b^2 \sin 	heta \cos 	heta = ax \sin 	heta - a^2 \sin 	heta \cos 	heta$. Rearranging terms,$ax \sin 	heta - by \cos 	heta = (a^2 - b^2) \sin 	heta \cos 	heta$. Dividing by $\sin 	heta \cos 	heta$,we get $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$.

The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(a \cos \theta, b \sin \theta)$ is

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