Find the equation

finding the slope of the tangent by differentiating the curve

$c^{2}\left(2 x+2 y \frac{d y}{d x}\right)=2 x y^{2}+2 x^{2} y \frac{d y}{d x}$

$2 x c^{2}-2 x y^{2}=2 x^{2} y \frac{d y}{d x}-2 y c^{2} \frac{d y}{d x}$

$\frac{d y}{d x}=\frac{x c^{2}-x y^{2}}{x^{2} y-y c^{2}}$

$\mathrm{m}$ (tangent) at $\left(\frac{\mathrm{c}}{\cos \theta}, \frac{\mathrm{c}}{\sin \theta}\right)=-\frac{\cos ^{2} \theta}{\sin ^{2} \theta}$

normal is perpendicular to tangent so, $m_{1} m_{2}=-1$

$\operatorname{m}($ normal $) \mathrm{at}\left(\frac{\mathrm{c}}{\cos \theta}, \frac{\mathrm{c}}{\sin \theta}\right)=\frac{\sin ^{3} \theta}{\cos ^{3} \theta}$

equation of tangent is given by $y-y_{1}=m($ tangent $)\left(x-x_{1}\right)$

$\mathrm{y}-\frac{\mathrm{c}}{\sin \theta}=-\frac{\cos ^{3} \theta}{\sin ^{3} \theta}\left(\mathrm{x}-\frac{\mathrm{c}}{\cos \theta}\right)$

equation of normal is given by $y-y_{1}=m($ normal $)\left(x-x_{1}\right)$

$y-\frac{c}{\sin \theta}=\frac{\sin ^{3} \theta}{\cos ^{3} \theta}\left(x-\frac{c}{\cos \theta}\right)$