Question

If the tangent drawn at point $(t^2,2t)$ 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:

• 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. $t=\pm \frac{1}{\sqrt{5}}$

Answer 1

Answer: (A), (D)

$\theta ={\cos }^{-1}(-\frac{1}{\sqrt{5}})$
$t=\pm \frac{1}{\sqrt{5}}$

The equation of the tangent at $(t^2,2t)$ to the parabola
$y^2=4x$ is
$2ty=2(x+t^2)$
$\Rightarrow ty=x+t^2$
$\Rightarrow x-ty+t^2=0$ (i)
The equation of the normal at point $(\sqrt{5}\cos \theta ,2\sin \theta )$ on the ellipse $5x^2+5y^2=20$ is
$\Rightarrow \left(\sqrt{5}\sec \theta \right)x-\left(2\cos ec\theta \right)y=5-4$
$\Rightarrow \left(\sqrt{5}\sec \theta \right)x-\left(2\cos ec\theta \right)y=1$ (ii)
Given that Eqs. (i) and (ii) represent the same line.
$\Rightarrow \frac{\sqrt{5}\sec \theta }{1}=\frac{-2\cos ec\theta }{-t}=\frac{-1}{t^2}$
$\Rightarrow t=\frac{2}{\sqrt{5}}\cot \theta$ and $t=-\frac{1}{2}\sin \theta$
$\Rightarrow \frac{2}{\sqrt{5}}\cot \theta =-\frac{1}{2}\sin \theta$
$\Rightarrow 4\cos \theta =-\sqrt{5}{\sin }^2\theta$
$\Rightarrow \sqrt{5}{\cos }^2\theta -4\cos \theta -\sqrt{5}=0$
$\Rightarrow (\cos \theta -\sqrt{5})(\sqrt{5}\cos \theta +1)=0$
$\Rightarrow \cos \theta =-\frac{1}{\sqrt{5}}[\because \cos \theta \ne \sqrt{5}]$
$\Rightarrow \theta ={\cos }^{-1}(-\frac{1}{\sqrt{5}})$
Putting $\cos \theta =-\frac{1}{\sqrt{5}}$ in $t=-\frac{1}{2}\sin \theta$, we get
$t=-\frac{1}{2}\sqrt{1-\frac{1}{5}}=\pm \frac{1}{\sqrt{5}}$
Hence, $\theta ={\cos }^{-1}(-\frac{1}{\sqrt{5}})$ and $t=\pm \frac{1}{\sqrt{5}}$.