Question

Find the angle of intersection between the curves $y^2=4x$ and $x^2+y^2=5$.

• A. ${\cos }^{-1}3$
• B. ${\sin }^{-1}3$
• C. ${\cot }^{-1}3$
• D. ${\tan }^{-1}3$

Answer 1

The correct option is D ${\tan }^{-1}3$

First we find the point of intersection of
$y^2=4x$ and $x^2+y^2=5$
Replacing $y^2=4x$

$x^2+4x=5$

$x^2+4x-5=0$
Solving we get
$x=1,-5$
For $x=1,y=\pm 2$
For $x=-5,$y is imaginary
So point of intersection are $(1,+2),(1,-2)$

For point $(1,2)$

Slope of $y^2=4x$

$m_1=\frac{dy}{dx}=\frac{4}{2y}=\frac{4}{2\times 2}=1$
Slope of $x^2+y^2=5$

$m_2=\frac{dy}{dx}=\frac{-2x}{2y}=\frac{-2\times 1}{2\times 2}=\frac{-1}{2}$
So the angle of intersection will be

$\theta ={\tan }^{-1}\left(\frac{m_1-m_2}{1+m_1{m}_2}\right)$

$\theta ={\tan }^{-1}\left(\frac{1-\frac{-1}{2}}{1+1\times \frac{-1}{2}}\right)$

$\theta ={\tan }^{-1}\left(\frac{\frac{3}{2}}{\frac{1}{2}}\right)$
$\theta ={\tan }^{-1}\left(3\right)$