Question

Find the angle if intersection of curves $y=4-x^2$ and $y=x^2$.

Answer 1

Consider the given equation .

$y=4-x^2......\left(1\right)$

$y=x^2........\left(2\right)$

On solving the equation ,we get

$x^2=4-x^2$

$2x^2=4$

$x=\pm \sqrt{2}$

Differentiate equation (1)with respect to x we get,

$\frac{dy}{dx}=-2x\therefore {\left(\frac{dy}{dx}\right)}_{\sqrt{2},2}=-2\sqrt{2}$= $m_1$

Differentiate equation (2) with respect to x we get,

$\frac{dy}{dx}=2x\therefore {\left(\frac{dy}{dx}\right)}_{\sqrt{2},2}=2\sqrt{2}$ $m_2$

The angle between the two curves is given as,$\tan \theta =\left|\frac{m_1-m_2}{1-m_1{m}_2}\right|$

Substitute the value of $m_1$ and $m_2$ ,we get

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

$=\frac{4\sqrt{2}}{7}$

Hence this is the answer.