Q108 / 160 · TS EAMCET 2023 (13 May Shift 2) · Mathematics

For $h, k \in N$, let $\mathrm{P}(h, k)$ 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 $P, \operatorname{then} \tan \theta=$

  • A
    $\frac{27}{11}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{\pi}{2}$
  • Correct answer
    D
    3

Solution

$\begin{aligned} & \text {} x^2 y-x^3=8 \\ & x^2 \cdot \frac{d y}{d x}+2 x y-3 x^2=0 \\ & \frac{d y}{d x}=\frac{3 x^2-2 x y}{x^2}=3-2\left(\frac{y}{x}\right)=m_1\end{aligned}$ $$ \begin{aligned} & y^3-x y^2=32 \\ & \Rightarrow 3 y^2 \frac{d y}{d x}-\left\{x \cdot 2 y \frac{d y}{d x}+y^2\right\}=0 \\ & \Rightarrow \frac{d y}{d x} \cdot\left(3 y^2-2 x y\right)-y^2=0 \\ & \Rightarrow \frac{d y}{d x}=\frac{y^2}{3 y^2-2 x y}=\frac{y}{3 y-2 x}=m_2 \end{aligned} $$ Also (2) $\div(1)$, $$ \begin{aligned} & \quad \frac{y^2(y-x)}{x^2(y-x)}=4 \Rightarrow \frac{y^2}{x^2}=4 \Rightarrow y=2 x \\ & \text { Putting } y=2 x \text { in (1) } \\ & \quad 2 x^3-x^3=8 \Rightarrow x=2 \\ & \therefore \quad y=4 \\ & \therefore \quad m_1=3-2(2)=-1 \text { and } m_2=\frac{4}{12-4}=\frac{1}{2} \\ & \therefore \quad \tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|=\left|\frac{\frac{1}{2}+1}{1-\frac{1}{2}}\right|=3 \end{aligned} $$

Hence, the correct answer is option: 'D'.

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