Question

The angle of intersection of $y^2=4x$ and $x^2=4y$ is ${90}^0$ and ${\tan }^{-1}\left(\frac{m}{n}\right)$ then $|m+n|$ is equal to ( $m$ and $n$ are coprime)

Answer 1

$y^2=4x2y\frac{dy}{dx}=4x\frac{dy}{dx}=\frac{2x}{y}\left(m_1\right)x^2=4y2x=4\frac{dy}{dx}\frac{dy}{dx}=\frac{x}{2}\left(m_2\right)y^2=4x{y}^4=16x^2{y}^4=16\times 4yy(y^3-64)=0,y=4x=4$