Q106 / 160 · AP EAMCET 2022 (08 Jul Shift 2) · Mathematics
If the angle between the curves $y^2=4 x$ and $y=e^{-x / 2}$ is $\theta$, then $\operatorname{cosec}^2(\theta / 2)=$
- Correct answerA$2$
- B$3$
- C$\sqrt{3}$
- D$\sqrt{2}$
Solution
The given curves are $y^2=4 x$ and $y=e^{-x / 2}$.
Let $\left(x_1, y_1\right)$ be the point of intersection of both curves.
Now, $y^2=4 x$
$\begin{array}{ll}\Rightarrow & \frac{d y}{d x}=\frac{2}{y} \\ \Rightarrow & m_1=\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=\frac{2}{y_1}\end{array}$
Also, $y=e^{-x / 2}$
$\begin{array}{ll}\Rightarrow & \frac{d y}{d x}=\left(-\frac{1}{2}\right) e^{-x / 2} \\ \Rightarrow & m_2=\left(\frac{d y}{d x}\right)_{\left(x_1, y_1\right)}=-\frac{1}{2} e^{-x_1 / 2}=-\frac{1}{2} y_1\end{array}$
Given that $\theta$ is the angle between the curves.
$\tan \theta=\left|\frac{m_2-m_1}{1+m_1 m_2}\right|$
$=\left|\frac{-\frac{y_1}{2}-\frac{2}{y_1}}{1+\left(-\frac{y_1}{2} \times \frac{2}{y_1}\right)}\right|=\infty$
$\therefore \quad \dot{\theta}=\pi / 2$
Now, $\operatorname{cosec}^2(\theta / 2)$
$=\operatorname{cosec}^2(\pi / 4)=(\sqrt{2})^2=2$
Hence, the correct answer is option: 'A'.