Question

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

• A. $Q\equiv (-1,\frac{4}{\sqrt{5}})$
• B. $Q\equiv (-1,-\frac{4}{\sqrt{5}})$
• C. $P\equiv (\frac{1}{5},-\frac{2}{\sqrt{5}})$
• D. $P\equiv (\frac{1}{5},\frac{2}{\sqrt{5}})$

Answer 1

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

The correct options are
A $Q\equiv (-1,\frac{4}{\sqrt{5}})$
B $Q\equiv (-1,-\frac{4}{\sqrt{5}})$
C $P\equiv (\frac{1}{5},-\frac{2}{\sqrt{5}})$
D $P\equiv (\frac{1}{5},\frac{2}{\sqrt{5}})$


The equation of the tangent at $P(t^2,2t)$ on the parabola $y^2=4x$ is :
$\Rightarrow x-ty+t^2=0\dots \left(1\right)$

Equation of the normal at point $Q(\sqrt{5}\cos \theta ,2\sin \theta )$ on the ellipse $4x^2+5y^2=20$ is :
$\left(\sqrt{5}\sec \theta \right)x-\left(2\text{cosec}\theta \right)y=5-4$
$\Rightarrow \left(\sqrt{5}\sec \theta \right)x-\left(2\text{cosec}\theta \right)y=1\dots \left(2\right)$

Given that equations $\left(1\right)$ and $\left(2\right)$ represent the same line.
$\Rightarrow \frac{\sqrt{5}\sec \theta }{1}=\frac{-2\text{cosec}\theta }{-t}=\frac{-1}{t^2}$
$\Rightarrow t=\frac{2}{\sqrt{5}}\cot \theta$ and $t=-\frac{1}{2}\sin \theta$
$\Rightarrow \frac{2}{\sqrt{5}}\cot \theta =-\frac{1}{2}\sin \theta$
$\Rightarrow 4\cos \theta =-\sqrt{5}{\sin }^2\theta$
$\Rightarrow 4\cos \theta =-\sqrt{5}(1-{\cos }^2\theta )$
$\Rightarrow \sqrt{5}{\cos }^2\theta -4\cos \theta -\sqrt{5}=0$
$\Rightarrow (\cos \theta -\sqrt{5})(\sqrt{5}\cos \theta +1)=0$
$\Rightarrow \cos \theta =-\frac{1}{\sqrt{5}}$ or $\cos \theta =\sqrt{5}$ (not possible)
$\Rightarrow \cos \theta =-\frac{1}{\sqrt{5}}\Rightarrow \sin \theta =\pm \frac{2}{\sqrt{5}}$
$\Rightarrow t=\mp \frac{1}{\sqrt{5}}$

Hence, $Q\equiv (-1,\pm \frac{4}{\sqrt{5}}),P\equiv (\frac{1}{5},\mp \frac{2}{\sqrt{5}})$