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(D) Given curves are $x^2 y - x^3 = 8$ $(1)$ and $y^3 - x y^2 = 32$ $(2)$.Differentiating $(1)$ with respect to $x$: $x^2 \frac{dy}{dx} + 2xy - 3x^2 =

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$3$

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(D) Given curves are $x^2 y - x^3 = 8$ $(1)$ and $y^3 - x y^2 = 32$ $(2)$.Differentiating $(1)$ with respect to $x$: $x^2 \frac{dy}{dx} + 2xy - 3x^2 = 0 \Rightarrow \frac{dy}{dx} = \frac{3x^2 - 2xy}{x^2} = 3 - 2(\frac{y}{x}) = m_1$.Differentiating $(2)$ with respect to $x$: $3y^2 \frac{dy}{dx} - (x \cdot 2y \frac{dy}{dx} + y^2) = 0 \Rightarrow \frac{dy}{dx}(3y^2 - 2xy) = y^2 \Rightarrow \frac{dy}{dx} = \frac{y^2}{3y^2 - 2xy} = \frac{y}{3y - 2x} = m_2$.Dividing $(2)$ by $(1)$: $\frac{y^2(y - x)}{x^2(y - x)} = \frac{32}{8} = 4 \Rightarrow \frac{y^2}{x^2} = 4 \Rightarrow y = 2x$ (since $h, k \in N$).Substituting $y = 2x$ into $(1)$: $x^2(2x) - x^3 = 8 \Rightarrow x^3 = 8 \Rightarrow x = 2$. Thus, $y = 4$.At $P(2, 4)$, $m_1 = 3 - 2(\frac{4}{2}) = 3 - 4 = -1$.At $P(2, 4)$, $m_2 = \frac{4}{3(4) - 2(2)} = \frac{4}{12 - 4} = \frac{4}{8} = \frac{1}{2}$.The angle $	heta$ between the curves is given by $	an 	heta = |\frac{m_2 - m_1}{1 + m_1 m_2}| = |\frac{1/2 - (-1)}{1 + (-1)(1/2)}| = |\frac{3/2}{1/2}| = 3$.

For $h, k \in N$, let $P(h, k)$ be the point of intersection of the curves $x^2 y - x^3 = 8$ and $y^3 - x y^2 = 32$. If $\theta$ is the acute angle between these two curves at $P$, then $\tan \theta =$

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