Question

Find the angle between the curves given below :
$y^2=4x,x^2+y^2=5$

Answer 1

We have $y^2=5$ and $x^2+y^2=5$

Let's find the intersection point of the two curves

Substituting the value of $y^2=4x$

We get $x^2+4x-5=0$

$⟹(x+5)(x-1)=0$

So, $x=-5$ or $x=1$

For $x=5,y=\pm 2$

$x=-5$ is neglected as $y$ have real value.

Slope of tangent of curve $y^2=4x$ is$m_1=\frac{dy}{dx}=\frac{4}{2y}{|}_{(1,2)}=1$

Slope of tangent of curve $x^2+y^2=5$ is$m_2=\frac{dy}{dx}=\frac{-x}{y}{|}_{(1,2)}=\frac{-1}{2}$

Let angle between them be $\theta$

Then $\tan \theta =|\frac{m_1-m_2}{1+m_1{m}_2}|=|\frac{1+\left)\right(1/2)}{1-\left(1/2\right)}|=3$

$⟹\theta ={\tan }^{-1}3$