If the tangent drawn to the parabola $y^2=4 x$ at $\left(t^2, 2 t\right)$ is the normal to the ellipse $4…

1. $5 t^4+4 t^2=1$
2. $\frac{5}{t^4}+\frac{100}{t^2}=1$
3. $t=\sin \theta$
4. $\cos \theta=t+1$

Solution

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 $

Asked in: AP EAMCET 2019 (21 Apr Shift 1)

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