Question

If the tangent drawn at a point $(t^2,2t)$ on the parabola $y^2=4x$ is same as normal drawn at $(\sqrt{5}\cos \alpha ,2\sin \alpha )$ on the ellipse $\frac{x^2}{5}+\frac{y^2}{4}=1$, then which of following is true.

• A. $t=\pm \frac{1}{\sqrt{5}}$
• B. $\alpha =-{\tan }^{-1}2$
• C. $\alpha ={\tan }^{-1}2$
• D. None of these

Answer 1

Answer: (A), (B), (C)

The correct options are
A $t=\pm \frac{1}{\sqrt{5}}$
B $\alpha =-{\tan }^{-1}2$
C $\alpha ={\tan }^{-1}2$

Equation of tangent to $y^2=4x$ at $(t^2,2t)$ is $x=ty-t^2$ ....(1)

Equation of normal to ellipse $\frac{x^2}{5}+\frac{y^2}{4}=1$ at $(\sqrt{5}\cos \alpha ,2\sin \alpha )$ is $\sqrt{5}\sec \alpha x-2y\csc \alpha =1$ ....(2)

Given (1) $=$ (2)

$\Rightarrow \sqrt{5}\sec \alpha =\frac{2\csc \alpha }{t}=-\frac{1}{t^2}\Rightarrow \cos \alpha =-\sqrt{5}{t}^2$ and $\sin \alpha =-2t$

$\Rightarrow {\cos }^2\alpha +{\sin }^2\alpha =5t^4+4t^2=1\Rightarrow t^2=\frac{1}{5}$

$\therefore$ (A) is true

and $\frac{\sin \alpha }{\cos \alpha }=-\frac{2t}{-\sqrt{5}{t}^2}=\frac{2}{\sqrt{5}}\times \frac{1}{t}=\frac{2}{\sqrt{5}}\times (\pm 5)$

$\therefore \tan \alpha =\pm 2$