Question

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

• A. $\theta ={\cos }^{-1}(-\frac{1}{\sqrt{5}})$
• B. $\theta ={\cos }^{-1}\left(\frac{1}{\sqrt{5}}\right)$
• C. $t=-\frac{2}{\sqrt{5}}$
• D. $t=-\frac{1}{\sqrt{5}}$

Answer 1

Answer: (A), (D)

The correct options are
A $\theta ={\cos }^{-1}(-\frac{1}{\sqrt{5}})$
D $t=-\frac{1}{\sqrt{5}}$

The equation of the tangent at $(t^2,2t)$ to the parabola $y^2=4x$ is $2ty=2(x+t^2)$
$\Rightarrow ty=x+t^2\Rightarrow x-ty+t^2=0...\left(1\right)$
The equation of the normal at $(\sqrt{5}\cos \theta ,2\sin \theta )$ on the ellipse $5x^2+5y^2=20$ is
$\left(\sqrt{5}\sec \theta \right)x-\left(2cosec\theta \right)y=5-4=1...\left(2\right)$
as $\left(1\right)$ and $\left(2\right)$ represent the same line,
$\frac{\sqrt{5}\sec \theta }{1}=\frac{-2cosec\theta }{-t}=\frac{-1}{t^2}$
$\Rightarrow t=\frac{2}{\sqrt{5}}\cot \theta$ and $t=-\frac{1}{2}\sin \theta$
$\frac{2}{\sqrt{5}}\cot \theta =-\frac{1}{2}\sin \theta$
$\Rightarrow 4\cos \theta =-\sqrt{5}{\sin }^2\theta$
$4\cos \theta =-\sqrt{5}(1-{\cos }^2\theta )$
$\sqrt{5}{\cos }^2\theta -4\cos \theta -\sqrt{5}=0$
$(\cos \theta -\sqrt{5})(\sqrt{5}\cos \theta +1)=0$

As $\cos \theta \ne \sqrt{5},\cos \theta =-\frac{1}{\sqrt{5}}$
$\Rightarrow \theta ={\cos }^{-1}(-\frac{1}{\sqrt{5}})$
$\Rightarrow t=-\frac{1}{2}\sqrt{1-\frac{1}{5}}=-\frac{1}{\sqrt{5}}$