Question

If $\theta$ is the acute angle of intersection at a real point of the circle $x^2+y^2=5$ and the parabola $y^2=4x$, then $\tan \theta$ is equal to

• A. $1$
• B. $\sqrt{3}$
• C. $3$
• D. $\frac{1}{\sqrt{3}}$

Answer 1

The correct option is C $3$

Given equations are
$x^2+y^2=5.....\left(i\right)$
and $y^2=4x.....\left(ii\right)$
On solving Eqs. (i) and (ii), we get
$x=-5,1$
at $x=-5,y^2=-20$ (imaginary value)
$\therefore$ at $x=1,y^2=4$
$\Rightarrow y=\pm 2$
Hence, point of intersection are $(1,2)$ and $(1,-2)$
On differentiating Eq. (i) w.r.t. x, we get
$2x+2y\frac{dy}{dx}=0$
$\Rightarrow \frac{dy}{dx}=-\frac{x}{y}$
$\therefore m_1={\left(\frac{dy}{dx}\right)}_{(1,2)}=-\frac{1}{2}$
And on differentiating Eq. (ii) w.r.t. x, we get
$2y\frac{dy}{dx}=4$
$\Rightarrow \frac{dy}{dx}=\frac{2}{y},m_2={\left(\frac{dy}{dx}\right)}_{(1,2)}=\frac{2}{2}=1$
Now, $\tan \theta =\left|\frac{m_1-m_2}{1+m_1{m}_2}\right|$
$=\left|\frac{-\frac{1}{2}-1}{1-\frac{1}{2}}\right|=\left|\frac{-\frac{3}{2}}{\frac{1}{2}}\right|=3$