If the equation of the normal to the curve x^ | vedclass

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(D) Given the curve $x^3 - y^2 = 0$.By differentiating with respect to $x$,we get $3x^2 - 2y \frac{dy}{dx} = 0$.So,$\frac{dy}{dx} = \frac{3x^2}{2y}$.A

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$2/9$

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(D) Given the curve $x^3 - y^2 = 0$.By differentiating with respect to $x$,we get $3x^2 - 2y \frac{dy}{dx} = 0$.So,$\frac{dy}{dx} = \frac{3x^2}{2y}$.At the point $(m^2, -m^3)$,the slope of the tangent is $\left( \frac{dy}{dx} ight)_{(m^2, -m^3)} = \frac{3(m^2)^2}{2(-m^3)} = \frac{3m^4}{-2m^3} = -\frac{3}{2}m$.The slope of the normal is $-\frac{1}{dy/dx} = -\frac{1}{-3m/2} = \frac{2}{3m}$.The equation of the normal at $(m^2, -m^3)$ is $y - (-m^3) = \frac{2}{3m}(x - m^2)$.$y + m^3 = \frac{2}{3m}x - \frac{2m^2}{3m} \Rightarrow y = \frac{2}{3m}x - \frac{2m}{3} - m^3$.Comparing this with the given equation $y = 3mx - 4m^3$,we equate the coefficients of $x$:$3m = \frac{2}{3m} \Rightarrow 9m^2 = 2 \Rightarrow m^2 = \frac{2}{9}$.Also,comparing the constant terms: $-4m^3 = -m^3 - \frac{2m}{3} \Rightarrow 3m^3 = \frac{2m}{3} \Rightarrow 9m^2 = 2 \Rightarrow m^2 = \frac{2}{9}$ (for $m 
eq 0$).

If the equation of the normal to the curve $x^3 - y^2 = 0$ at the point $(m^2, -m^3)$ is $y = 3mx - 4m^3$,then $m^2 = \dots\dots$.

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