Question

If the tangent drawn at point $\left(t^2, 2 t\right)$ on the parabola $y^2=4 x$ is same as the normal drawn at point $(\sqrt{5} \cos \theta, 2 \sin \theta)$ on the ellipse $4 x^2+5 y^2=20$. Then which of the following is/are possible?

• A. $\theta=\cos ^{-1}\left(-\frac{1}{\sqrt{5}}\right)$
• 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 equation of the tangent at $\left(t^2, 2 t\right)$ to the parabola $y^2=4 x$ is
$ 2 t y=2\left(x \mid t^2\right) \Rightarrow t y=x \mid t^2$
$\Rightarrow x-t y+t^2=0$
The equation of the normal at point $(\sqrt{5} \cos \theta, 2 \sin \theta)$ on the ellipse $4 x^2+5 y^2-20$ is
$\Rightarrow (\sqrt{5} \sec \theta) x (2 \operatorname{cosec} \theta) y=5 4 $
$\rightarrow (\sqrt{5} \sec 0) x-(2 \operatorname{cosec} 0) y=1$
Given that equations (i) and (ii) represent the same line.
$\Rightarrow \frac{\sqrt{5} \sec \theta}{1}=\frac{-2 \operatorname{cosec} \theta}{-t}=\frac{-1}{t^2} \Rightarrow l=\frac{2}{\sqrt{5}} \cot \theta \text { and } t=\frac{1}{2} \sin \theta$
$\Rightarrow \sqrt{5}_2^2 \cot \theta={ }_2^1 \sin \theta \Rightarrow 4 \cos \theta=-\sqrt{5} \sin ^2 \theta$
$\Rightarrow 4 \cos \theta=-\sqrt{5}\left(1-\cos ^2 \theta\right) \Rightarrow \sqrt{5} \cos ^2 \theta-4 \cos \theta-\sqrt{5}=0$
$\Rightarrow (\cos 0-\sqrt{5})(\sqrt{5} \cos 0+1)=0 \Rightarrow \cos \theta=-\frac{1}{\sqrt{5}} [\because \cos 0 \neq \sqrt{5}]$
$\Rightarrow \theta=\cos ^{-1}\left(-\frac{1}{\sqrt{5}}\right)$
Putting $\cos \theta--\frac{1}{\sqrt{5}}$ in $t--\frac{1}{2} \sin \theta$, we get : $t--\frac{1}{2} \sqrt{1-\frac{1}{5}}--\frac{1}{\sqrt{5}}$
Hence, $\theta=\cos ^{-1}\left(-\frac{1}{\sqrt{5}}\right)$ and $t=-\frac{1}{\sqrt{5}}$.