Two straight lines one being a tangent to $y^2=4ax$ and the other to $x^2=4$ by are at right angles. The locus of their point of intersection is given by :

Let P(h, k) be the point of intersection equation of tangent to $y^2=4ax$ is also, equation of tangent to $x^2=4by$ is

$y=m_{2}x-b{m}_2^2=\frac{1}{m_1}\times \frac{b}{m_1^2}\dots ..\left(2\right)$

$(\because m_1{m}_2=-1)$ since P(h, k) is their point of intersection We have by substituting P(h, k) in (1) and (2)

$\frac{m_1^2}{-bk}=\frac{m_1}{ak-bh}=\frac{1}{b^2+k^2}$

We get

$(h^2+k^2)(ah+bk)+(ak-bh{)}^2=0\Rightarrow$$(x^2+y^2)(ax+by)+(ay+bx{)}^2=0$