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AP EAPCET 08-Jul-2022 Shift 2 Paper

Section: Mathematics

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Question : 66 of 160

Marks: +1, -0

If the angle between the curves

y^2=4x

y=e^{\frac{-x}{2}}

\theta

\csc^{2}\left(\frac{\theta}{2}\right)=

$2$
$3$
$\sqrt{3}$
$\sqrt{2}$

Solution: 👈: Video Solution

The given curves are

y^2=4x

and

y=e^{\frac{-x}{2}}

.

Let

(x_1, y_1)

be the point of intersection of both curves.

Now,

y^2=4x

\Rightarrow \frac{dy}{dx}=\frac{2}{y}

\Rightarrow m_1= \left(\frac{dy}{dx}\right)_{(x_1, y_1)}=\frac{2}{y_1}

Also,

y=e^{\frac{-x}{2}}

\Rightarrow \frac{dy}{dx}= \left(-\frac{1}{2}\right) e^{\frac{-x}{2}}

\Rightarrow m_2= \left(\frac{dy}{dx}\right)_{(x_1, y_1)}=-\frac{1}{2} e^{\frac{-x_1}{2}}=-\frac{1}{2} y_1

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 \theta = \frac{\pi}{2}

Now,

\csc^{2}\left(\frac{\theta}{2}\right)

=\csc^{2}\left(\frac{\pi}{4}\right)=\left(\sqrt{2}\right)^2=2

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