Question

If the tangent drawn at point $\left(t^2,2t\right)$ on the parabola $y^2=4x$ is same as the normal drawn at point $(\sqrt{5}\cos \theta ,2\sin \theta )$ on the ellipse $4x^2+5y^2=20$. 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 $\left(t^2,2t\right)$ to the parabola $y^2=4x$ is
$2ty=2\left(x\mid t^2\right)\Rightarrow ty=x\mid t^2$
$\Rightarrow x-ty+t^2=0$
The equation of the normal at point $(\sqrt{5}\cos \theta ,2\sin \theta )$ on the ellipse $4x^2+5y^2-20$ is
$\Rightarrow (\sqrt{5}\sec \theta )x(2\mathrm{cosec}\theta )y=54$
$\to (\sqrt{5}\sec 0)x-(2\mathrm{cosec}0)y=1$
Given that equations (i) and (ii) represent the same line.
$\Rightarrow \frac{\sqrt{5}\sec \theta }{1}=\frac{-2\mathrm{cosec}\theta }{-t}=\frac{-1}{t^2}\Rightarrow l=\frac{2}{\sqrt{5}}\cot \theta \text{ and }t=\frac{1}{2}\sin \theta$
$\Rightarrow {\sqrt{5}}_2^2\cot \theta ={}_2^1\sin \theta \Rightarrow 4\cos \theta =-\sqrt{5}{\sin }^2\theta$
$\Rightarrow 4\cos \theta =-\sqrt{5}\left(1-{\cos }^2\theta \right)\Rightarrow \sqrt{5}{\cos }^2\theta -4\cos \theta -\sqrt{5}=0$
$\Rightarrow (\cos 0-\sqrt{5})(\sqrt{5}\cos 0+1)=0\Rightarrow \cos \theta =-\frac{1}{\sqrt{5}}[\because \cos 0\ne \sqrt{5}]$
$\Rightarrow \theta ={\cos }^{-1}\left(-\frac{1}{\sqrt{5}}\right)$
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 $t=-\frac{1}{\sqrt{5}}$.