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# Find the equation

Question:

Find the equation of the tangent and the normal to the following curves at the indicated points:

$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $(\sqrt{2} a, b)$

Solution:

finding slope of the tangent by differentiating the curve

$\frac{x}{a^{2}}-\frac{y}{b^{2}} \frac{d y}{d x}=0$

$\frac{d y}{d x}=\frac{x b^{2}}{y a^{2}}$

$\mathrm{m}(\operatorname{tangent})$ at $(\sqrt{2} \mathrm{a}, \mathrm{b})=\frac{\sqrt{2} \mathrm{ab}^{2}}{\mathrm{ba}^{2}}$

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

$\mathrm{m}$ (normal) at $(\sqrt{2} \mathrm{a}, \mathrm{b})=-\frac{\mathrm{ba}^{2}}{\sqrt{2} \mathrm{ab}^{2}}$

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

$y-b=\frac{\sqrt{2} a b^{2}}{b a^{2}}(x-\sqrt{2} a)$

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

$y-b=-\frac{b a^{2}}{\sqrt{2} a b^{2}}(x-\sqrt{2} a)$