If the tangent to the curve y^{2} = x^{3} at | vedclass

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(D) Given the curve $y^{2} = x^{3}$. Differentiating with respect to $x$, we get $2y \frac{dy}{dx} = 3x^{2}$, so $\frac{dy}{dx} = \frac{3x^{2}}{2y}$.A

Vedclass · Accepted Answer

$-\frac{4}{9}$

Vedclass · Answer

(D) Given the curve $y^{2} = x^{3}$. Differentiating with respect to $x$, we get $2y \frac{dy}{dx} = 3x^{2}$, so $\frac{dy}{dx} = \frac{3x^{2}}{2y}$.At the point $(m^{2}, m^{3})$, the slope of the tangent $T_{1}$ is $\frac{3(m^{2})^{2}}{2(m^{3})} = \frac{3m^{4}}{2m^{3}} = \frac{3m}{2}$.The equation of the tangent at $(m^{2}, m^{3})$ is $y - m^{3} = \frac{3m}{2}(x - m^{2})$, which simplifies to $2y - 2m^{3} = 3mx - 3m^{3}$, or $3mx - 2y - m^{3} = 0$.At the point $(M^{2}, M^{3})$, the slope of the tangent is $\frac{3M}{2}$. Thus, the slope of the normal $N_{2}$ is $-\frac{2}{3M}$.The equation of the normal at $(M^{2}, M^{3})$ is $y - M^{3} = -\frac{2}{3M}(x - M^{2})$, which simplifies to $3My - 3M^{4} = -2x + 2M^{2}$, or $2x + 3My - (3M^{4} + 2M^{2}) = 0$.Since the tangent at $(m^{2}, m^{3})$ is the same as the normal at $(M^{2}, M^{3})$, the coefficients must be proportional:$\frac{3m}{2} = \frac{-2}{3M} \Rightarrow 9mM = -4 \Rightarrow mM = -\frac{4}{9}$.Also, the constant terms must satisfy the ratio: $\frac{-m^{3}}{-(3M^{4} + 2M^{2})} = \frac{3m}{2} \Rightarrow 2m^{3} = 3m(3M^{4} + 2M^{2}) \Rightarrow 2m^{2} = 9M^{4} + 6M^{2}$.Substituting $m = -\frac{4}{9M}$, we get $2(-\frac{4}{9M})^{2} = 9M^{4} + 6M^{2} \Rightarrow \frac{32}{81M^{2}} = 9M^{4} + 6M^{2} \Rightarrow 32 = 729M^{6} + 486M^{4}$.This confirms the consistency of the slope condition $mM = -\frac{4}{9}$.

If the tangent to the curve $y^{2} = x^{3}$ at $(m^{2}, m^{3})$ is also a normal to the curve at $(M^{2}, M^{3})$, then the value of $mM$ is

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