If the tangent drawn at  the point (t², 2t) on the parabola y² = 4x is the same as the normal drawn at the point $(\sqrt{5}\cos \mathrm{\theta },2\sin \mathrm{\theta })$ on the ellipse 4x² + 5y² = 20 then

Tangent at t to ${\mathrm{y}}^2=4\mathrm{x}\text{ is }\mathrm{x}-\mathrm{ty}+{\mathrm{t}}^2=0$....... (1)
normal at $(\sqrt{5}\cos \mathrm{\theta },2\sin \mathrm{\theta })$ to  the ellipse 4x² + 5y² = 20 is

 $\begin{array}{l}\Rightarrow (\sqrt{5}\sec \mathrm{\theta })\mathrm{x}-(2\mathrm{cosec}\mathrm{\theta })\mathrm{y}=1.... \left(2\right), \\ Since \left(1\right) and \left(2\right) represents the same line then\\ \frac{\sqrt{5}\sec \mathrm{\theta }}{1}=\frac{-2\mathrm{cosec}\mathrm{\theta }}{-t}=\frac{-1}{t^2}\\ t=\frac{2}{\sqrt{5}}cot\theta and t=\frac{-1}{2}\sin \theta \\ \Rightarrow \frac{2}{\sqrt{5}}cot\theta =\frac{-1}{2}\sin \theta \\ \Rightarrow \sqrt{5}{\cos }^2\theta -4\cos \theta -\sqrt{5}=0\\ \Rightarrow (\sqrt{5}\cos \theta +1)(\cos \theta -\sqrt{5})=0\\ \Rightarrow \cos \theta =\frac{-1}{\sqrt{5}}(\because \cos \theta \ne \sqrt{5})\\ \Rightarrow \mathrm{\theta }={\cos }^{-1}\left(-\frac{1}{\sqrt{5}}\right)\\ sub \cos \theta value in t we get \\ \Rightarrow \mathrm{t}=-\frac{1}{\sqrt{5}}\end{array}$