Find the equation of the normal at the point | vedclass

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(A) The equation of the given curve is $a y^{2}=x^{3}$.Differentiating both sides with respect to $x$,we get:$2 a y \frac{d y}{d x} = 3 x^{2}$$\frac{d

Vedclass · Accepted Answer

$2 x+3 m y-a m^{2}(2+3 m^{2})=0$

Vedclass · Answer

(A) The equation of the given curve is $a y^{2}=x^{3}$.Differentiating both sides with respect to $x$,we get:$2 a y \frac{d y}{d x} = 3 x^{2}$$\frac{d y}{d x} = \frac{3 x^{2}}{2 a y}$.The slope of the tangent at the point $(a m^{2}, a m^{3})$ is:$m_{t} = \left. \frac{d y}{d x} \right|_{(a m^{2}, a m^{3})} = \frac{3(a m^{2})^{2}}{2 a(a m^{3})} = \frac{3 a^{2} m^{4}}{2 a^{2} m^{3}} = \frac{3 m}{2}$.The slope of the normal at $(a m^{2}, a m^{3})$ is $m_{n} = -\frac{1}{m_{t}} = -\frac{2}{3 m}$.The equation of the normal at $(a m^{2}, a m^{3})$ is given by:$y - a m^{3} = -\frac{2}{3 m}(x - a m^{2})$$3 m y - 3 a m^{4} = -2 x + 2 a m^{2}$$2 x + 3 m y - 3 a m^{4} - 2 a m^{2} = 0$$2 x + 3 m y - a m^{2}(2 + 3 m^{2}) = 0$.

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

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