Find the equation


Find the equation of the normal to the curve $a y^{2}=x^{3}$ at the point $\left(a m^{2}, a m^{3}\right)$.


finding the slope of the tangent by differentiating the curve

$2 a y \frac{d y}{d x}=3 x^{2}$

$\frac{d y}{d x}=\frac{3 x^{2}}{2 a y}$

$\mathrm{m}$ (tangent) at $\left(\mathrm{am}^{2}, \mathrm{am}^{3}\right)$ is $\frac{3 \mathrm{~m}}{2}$

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

$\mathrm{m}$ (normal) at $\left(\mathrm{am}^{2}, \mathrm{am}^{3}\right)$ is $-\frac{2}{3 \mathrm{~m}}$

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

$y-a m^{3}=-\frac{2}{3 m}\left(x-a m^{2}\right)$

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