Given, tangent to parabola $y^2=4 x$ at $\left(t^2, 2 t\right)$ is
$$
y \cdot 2 t=2\left(x+t^2\right) \Rightarrow y t=x+t^2
$$


Normal to the ellipse $4 x^2+5 y^2=20$
or $\frac{x^2}{5}+\frac{y^2}{4}=1$ at $(\sqrt{5} \cos \theta, 2 \sin \theta)$
$\Rightarrow$ Slope of normal $=5 y / 4 x=\frac{\sqrt{5}}{2} \tan \theta$
$\therefore$ Equation of line is
$$
y-2 \sin \theta=\frac{\sqrt{5}}{2} \tan \theta(x-\sqrt{5} \cos \theta)
$$

By comparing Eqs. (i) and (ii), we get
$$
t=-\frac{\sin \theta}{2}
$$

$\begin{aligned} \frac{1}{t} & =\frac{\sqrt{5}}{2} \tan \theta \\ \Rightarrow \quad \tan \theta & =\frac{2}{\sqrt{5 t}}\end{aligned}$

From Eqs. (iii) and (iv), we get
$$
-2 t=\frac{2}{\sqrt{4+5 t^2}}
$$

On squaring both sides, we get
$$
t^2\left(4+5 t^2\right)=1 \Rightarrow 5 t^4+4 t^2=1
$$