If $\mathrm{L}$ and ${\mathrm{L}}^1$ are ends of latusrectum of the parabola $9{\mathrm{y}}^2=4\mathrm{x}$ then the combined equation of $\mathrm{OL}$ and ${\mathrm{OL}}^1$ is

$$\begin{gathered} 9{\mathrm{y}}^2=4\mathrm{x} \\ {\mathrm{y}}^2=\frac{4}{9}\mathrm{x} \\ 4\mathrm{a}=4/9 \\ \mathrm{a}=1/9 \end{gathered}$$

ends of latus rectum are

$$\begin{gathered} \mathrm{L}=\left(\mathrm{a},2\mathrm{a}\right)=\left(\frac{1}{9},\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\right) \\ {\mathrm{L}}^1=\left(a,-2\mathrm{a}\right)=\left(\frac{1}{9},\raisebox{1ex}{$-2$}\!\left/ \!\raisebox{-1ex}{$9$}\right.\right) \\ \mathrm{Slope} \mathrm{of} \overline{OL}=2 \\ \mathrm{Eq}uation \mathrm{of} \overline{OL } \\ \mathrm{y}-0=2\left(\mathrm{x}-0\right) \\ 2\mathrm{x}-\mathrm{y}=0 \end{gathered}$$

 $$\begin{gathered} Slope of {\mathrm{OL}}^1=-2 \\ \mathrm{Eq}uation \mathrm{of} \overline{OL\text{'}} \\ \mathrm{y}-0=-2(\mathrm{x}-0) \\ 2\mathrm{x}+\mathrm{y}=0 \\ Their \mathrm{combined} \mathrm{eq}uation \mathrm{is} (2x-y)(2x+y)=0 \\ \Rightarrow 4{\mathrm{x}}^2-{\mathrm{y}}^2=0 \\ \Rightarrow 4{\mathrm{x}}^2={\mathrm{y}}^2 \end{gathered}$$