Question

The angle between $y^2=4x$ and $x^2+y^2=12$ at a point of their intersection is

• A. $ta{n}^{-1}\sqrt{2}$
• B. $ta{n}^{-1}2$
• C. $ta{n}^{-1}2\sqrt{2}$
• D. $ta{n}^{-1}(\frac{1}{2})$

Answer 1

Answer: (C)

Given equation of curve is
$y^2=4x$...(i)
and $x^2+y^2=12$...(ii)
Let the slope of curve (i) is $m_1$ and curve (ii) is $\left(m_2\right)$
Then, from Eqs. (i) and (ii),
$2y\frac{dy}{dx}=4\Rightarrow m_1=\frac{2}{y}$
and $2x+2y\frac{dy}{dx}=0$
$\Rightarrow m_2=\frac{-x}{y}$
Let the angle between curve at intersection point is ' $\theta$ '.
$\tan \theta =\left|\frac{m_1-m_2}{1+m_1\cdot m_2}\right|=\left|\frac{2/y-(-x/y)}{1+(2/y)(-x/y)}\right|$
$\tan \theta =\left|\frac{\frac{2}{y}+\frac{x}{y}}{1-\frac{2x}{y^2}}\right|=\left|\frac{(2+x)y}{y^2-2x}\right|$ (iii)
On solving Eqs. (i) and (ii), we get
$x^2+4x=12$
$\Rightarrow x^2+4x-12=0$
$\Rightarrow x^2+6x-2x-12=0$
$\Rightarrow (x+6)(x-2)=0$
$\Rightarrow x=-6,2($ here $x\ne -6)$
$\Rightarrow y=\pm 2\sqrt{2}$
So, the intersection point $=(2,\pm 2\sqrt{2})$
From Eq. (iii)
$\tan \theta =\left|\frac{(2+2)(\pm 2\sqrt{2})}{8-4}\right|=\left|4\frac{(\pm 2\sqrt{2})}{4}\right|$
$\tan \theta =|\pm 2\sqrt{2}|=2\sqrt{2}$
$\Rightarrow \theta ={\tan }^{-1}(2\sqrt{2})$