If from a point A, the two tangents drawn to the parabola ${\mathrm{y}}^2=4\mathrm{ax}$ are normals to the parabola ${\mathrm{x}}^2=4\mathrm{by}$, then

$$\begin{gathered} \mathrm{Given} \mathrm{parabolas} {\mathrm{y}}^2=4\mathrm{ax} -\left(1\right) and {\mathrm{x}}^2=4\mathrm{by} - \left(2\right) \\ \mathrm{Tange}nt \mathrm{of} \left(1\right) \mathrm{is} \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}} \\ \mathrm{it} \mathrm{is} \mathrm{normal} \mathrm{of} \left(2\right) \mathrm{then} \mathrm{condition} \mathrm{is} \\ \overline{)\mathrm{C}=2\mathrm{b}+\frac{\mathrm{b}}{{\mathrm{m}}^2} \\ \Rightarrow \frac{\mathrm{a}}{\mathrm{m}}=2\mathrm{b}+\frac{\mathrm{b}}{{\mathrm{m}}^2} \\ \Rightarrow \mathrm{am}=2{\mathrm{bm}}^2+\mathrm{b}} \\ \begin{array}{c}\Rightarrow 2{\mathrm{bm}}^2-\mathrm{am}+\mathrm{b}=0\\ \\ \\ Since ∆>0 then \\ {\mathrm{a}}^2>8{\mathrm{b}}^2\end{array} \end{gathered}$$