For what values of 'a' will the tangents drawn to parabola y² = 4ax from a point , not on the y axis , will be normal to the parabola x² = 4y

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$$\begin{gathered} \mathrm{We} \mathrm{know} \mathrm{that} \\ \mathrm{Equation} \mathrm{of} \mathrm{tangent} \mathrm{to} \mathrm{parabola} {\mathrm{y}}^2=4\mathrm{ax} \mathrm{is} \\ \mathrm{y}=\mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}} ;\mathrm{equation}\left(\mathrm{i}\right) \mathrm{where} \text{'}\mathrm{m}\text{'} \mathrm{is} \mathrm{slope} \mathrm{of} \mathrm{tangent}. \\ \mathrm{Also} \\ \mathrm{Equation} \mathrm{of} \mathrm{normal} \mathrm{to} \mathrm{parabola} {\mathrm{x}}^2=4\mathrm{ay} \mathrm{is}; \\ \mathrm{y}=\mathrm{mx}+2\mathrm{a}+\frac{\mathrm{a}}{{\mathrm{m}}^2}, \mathrm{where} \text{'}\mathrm{m}\text{'} \mathrm{is} \mathrm{slope} \mathrm{of} \mathrm{normal}. \\ \mathrm{In} \mathrm{given} \mathrm{question} \mathrm{we} \mathrm{have} {\mathrm{x}}^2=4\mathrm{y}. \\ \Rightarrow \mathrm{a}=1 \\ \Rightarrow \mathrm{y}=\mathrm{mx}+2+\frac{1}{{\mathrm{m}}^2} ;\mathrm{equation}\left(\mathrm{ii}\right) \\ \mathrm{According} \mathrm{to} \mathrm{question}, \mathrm{equation}\left(\mathrm{i}\right) \mathrm{and} \mathrm{equation}\left(\mathrm{ii}\right) \mathrm{should} \mathrm{be} \mathrm{identical}. \\ \mathrm{mx}+\frac{\mathrm{a}}{\mathrm{m}}=\mathrm{mx}+2+\frac{1}{{\mathrm{m}}^2} \\ \frac{\mathrm{a}}{\mathrm{m}}=2+\frac{1}{{\mathrm{m}}^2} \\ \Rightarrow \frac{\mathrm{a}}{\mathrm{m}}=\frac{2{\mathrm{m}}^2+1}{{\mathrm{m}}^2} \\ \Rightarrow \mathrm{am}=2{\mathrm{m}}^2+1 \\ \Rightarrow 2{\mathrm{m}}^2-\mathrm{am}+1=0 \\ \mathrm{For} \text{'}\mathrm{m}\text{'} \mathrm{to} \mathrm{be} \mathrm{real} \mathrm{and} \mathrm{distinct} \\ \mathrm{Discriminant}>0 \\ \Rightarrow {\left(-\mathrm{a}\right)}^2-4\times 2\times 1>0 \\ \Rightarrow {\mathrm{a}}^2-8>0 \\ \Rightarrow {\mathrm{a}}^2-{\left(2\sqrt{2}\right)}^2>0 \\ \Rightarrow \left(\mathrm{a}-2\sqrt{2}\right)\left(\mathrm{a}+2\sqrt{2}\right)>0 \\ \mathbf{\Rightarrow }\mathbf{a}\mathbf{\in }\left(\mathbf{-}\mathbf{\infty }\mathbf{,}\mathbf{-}\mathbf{ }\mathbf{2}\sqrt{\mathbf{2}}\right)\mathbf{\cup }\left(\mathbf{2}\sqrt{\mathbf{2}}\mathbf{,}\mathbf{ }\mathbf{\infty }\right)\mathbf{ }\left(\mathbf{Answer}\right) \end{gathered}$$

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