Question

Consider the curve $x=1-3t^2,y=t-3t^3$. The tangent at any point on the curve is inclined at an angle $\theta$ with the positive x-axis. If tangent at point $P(-2,2)$ cuts the curve again at point $Q$, then

• A. $\tan \theta \pm \sec \theta =3t$
• B. Coordinates of $Q$ is $(-\frac{1}{3},-\frac{2}{3})$
• C. Angle between tangents at $P$ and $Q$ is $\frac{\pi }{2}$
• D. Angle between tangents at $P$ and $Q$ is $\frac{\pi }{4}$

Answer 1

Answer: (A), (C)

The correct options are
A $\tan \theta \pm \sec \theta =3t$
C Angle between tangents at $P$ and $Q$ is $\frac{\pi }{2}$

$\frac{dy}{dx}=\frac{1-9t^2}{-6t}=\tan \theta$
$\Rightarrow 9t^2-6t\tan \theta -1=0$
$\Rightarrow 3t=\tan \theta \pm \sec \theta$

$P\equiv (-2,2)\Rightarrow t=-1$
$\Rightarrow \frac{dy}{dx}{|}_{t=-1}=-\frac{4}{3}$
Equation of tangent at $P$ is $y-2=-\frac{4}{3}(x+2)$
$\Rightarrow t-3t^3-2=-\frac{4}{3}(1-3t^2+2)$
$\Rightarrow 9t^3+12t^2-3t-6=0$
$\Rightarrow 3t^3+4t^2-t-2=0$
$\Rightarrow (t+1)(3t^2+t-2)=0$
$\Rightarrow (3t-2)(t+1{)}^2=0$
$\Rightarrow t=\frac{2}{3}$
$\therefore Q\equiv (-\frac{1}{3},-\frac{2}{9})$

$\frac{dy}{dx}{|}_{t=2/3}=\frac{3}{4}$
$m_P\times m_Q=-1$
Therefore, angle between tangents at $P$ and $Q$ is ${90}^{\circ }$