Q107 / 160 · AP EAMCET 2008 · Mathematics
The angle between the curves $y^2=4 x+4$ and $y^2=36(9-x)$ is
- A$30^{\circ}$
- B$45^{\circ}$
- C$60^{\circ}$
- Correct answerD$90^{\circ}$
Solution
Given curves are
$$
y^2=4 x+4 \text { and } y^2=36(9-x)
$$
On solving, we get the points $(8,6)$ and $(8,-6)$.
On differentiating Eq. (i), we get
$$
\begin{aligned}
& 2 y \frac{d y}{d x}=4 \text { and } 2 y \frac{d y}{d x}=-36 \\
\Rightarrow \quad & \frac{d y}{d x}=\frac{2}{y} \text { and } \frac{d y}{d x}=\frac{-18}{y}
\end{aligned}
$$
At point $(8,6)$,
$$
\begin{aligned}
& m_1=\frac{d y}{d x}=\frac{2}{6}=\frac{1}{3} \text { and } m_2=\frac{d y}{d x}=\frac{-18}{6}=-3 \\
& \therefore \quad \tan \theta=\frac{\frac{1}{3}+3}{1+\frac{1}{3} \times(-3)}=\frac{10 / 3}{0}=\infty \\
& \Rightarrow \quad \theta=\frac{\pi}{2}
\end{aligned}
$$
And at point $(8,-6)$,
$$
\begin{aligned}
& m_1=\frac{d y}{d x}=\frac{2}{-6}=-\frac{1}{3} \text { and } m_2=\frac{d y}{d x}=\frac{-18}{-6}=3 \\
& \therefore \quad \tan \theta=\frac{\left|-\frac{1}{3}-3\right|}{1+\left(-\frac{1}{3}\right) \cdot 3}=\frac{10 / 3}{0}=\infty \\
& \Rightarrow \quad \theta=\frac{\pi}{2}
\end{aligned}
$$
Hence, the correct answer is option: 'D'.