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 $\left(x_{1}, y_{1}\right)$


finding the slope of the tangent by differentiating the curve

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

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

$\mathrm{m}($ tangent $)$ at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=\frac{\mathrm{b}^{2} \mathrm{x}_{1}}{\mathrm{y}_{1} \mathrm{a}^{2}}$

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

$\mathrm{m}$ (normal) at $\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=-\frac{\mathrm{a}^{2} \mathrm{y}_{1}}{\mathrm{x}_{1} \mathrm{~b}^{2}}$

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

$\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{b}^{2} \mathrm{x}_{1}}{\mathrm{y}_{1} \mathrm{a}^{2}}\left(\mathrm{x}-\mathrm{x}_{1}\right)$

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

$y-y_{1}=-\frac{a^{2} y_{1}}{x_{1} b^{2}}\left(x-x_{1}\right)$

Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now