a double ordinate of the parabola y²=4ax is of length 8a. prove that the lines joining the vertex to its two ends are at right angles.

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$$\begin{gathered} \mathrm{Double} \mathrm{ordinate} \mathrm{is} \mathrm{perpendicular} \mathrm{to} \mathrm{axis} \mathrm{of} \mathrm{parabola}, \mathrm{hence} \mathrm{let} \mathrm{end} \mathrm{points} \\ \mathrm{of} \mathrm{double} \mathrm{ordinate} \mathrm{be} \\ \mathrm{P}=\left({\mathrm{at}}^2,2\mathrm{at}\right) \mathrm{and} \mathrm{P}\text{'}=\left({\mathrm{at}}^2,-2\mathrm{at}\right) \\ \mathrm{PP}\text{'}=8\mathrm{a} \\ \sqrt{{\left({\mathrm{at}}^2-{\mathrm{at}}^2\right)}^2+{\left(2\mathrm{at}+2\mathrm{at}\right)}^2}=8\mathrm{a} \\ \Rightarrow \sqrt{{\left(4\mathrm{at}\right)}^2}=8\mathrm{a} \\ \Rightarrow 4\mathrm{at}=8\mathrm{a} \\ \Rightarrow \mathrm{t}=2 \\ \mathrm{Hence} \mathrm{P}=\left({\mathrm{at}}^2,2\mathrm{at}\right)=\left(4\mathrm{a},4\mathrm{a}\right) \\ \mathrm{P}\text{'}=\left({\mathrm{at}}^2,-2\mathrm{at}\right)=\left(4\mathrm{a},-4\mathrm{a}\right) \\ \mathrm{Vertex} \mathrm{is} \mathrm{O}\left(0,0\right) \\ \mathrm{Slope} \mathrm{of} \mathrm{OP}={\mathrm{m}}_1=\frac{4\mathrm{a}-0}{4\mathrm{a}-0}=1 \\ \mathrm{Slope} \mathrm{of} \mathrm{OP}\text{'}={\mathrm{m}}_2=\frac{4\mathrm{a}-0}{-4\mathrm{a}-0}=-1 \\ \Rightarrow {\mathrm{m}}_1.{\mathrm{m}}_2=1\times \left(-1\right) \\ \Rightarrow {\mathrm{m}}_1.{\mathrm{m}}_2=-1 \\ \mathrm{Hence} \angle \mathrm{POP}\text{'}=\frac{\mathrm{\pi }}{2} \\ \mathbf{\therefore }\mathbf{ }\mathbf{ }\mathbf{Lines}\mathbf{ }\mathbf{joining}\mathbf{ }\mathbf{the}\mathbf{ }\mathbf{vertex}\mathbf{ }\mathbf{to}\mathbf{ }\mathbf{two}\mathbf{ }\mathbf{ends}\mathbf{ }\mathbf{of}\mathbf{ }\mathbf{double}\mathbf{ }\mathbf{ordinate}\mathbf{ }\mathbf{are}\mathbf{ }\mathbf{at}\mathbf{ }\mathbf{right}\mathbf{ }\mathbf{angles}\mathbf{.} \end{gathered}$$

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