Deepak Scored 45->99%ile with Bounce Back Crack Course. You can do it too!

Find the equation


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 $(\sqrt{2} a, b)$


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 $)\left(x-x_{1}\right)$

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

Leave a comment

Free Study Material