Question

Consider the curve $x=1-3t^2,y=t-3t^3,$ Let P(-2, 2) be a point on the curve. Let the tangent at P
cuts the curve again at Q. Then answer the following
The Point Q will be

• A. $(-2,-2)$
• B. $(-\frac{1}{2},-\frac{2}{9})$
• C. $(\frac{2}{3},\frac{2}{9})$
• D. $(-2,1)$

Answer 1

The correct option is B $(-\frac{1}{2},-\frac{2}{9})$

$x=1-3t^2,y=t-3t^{3}P=(-2,2)\Rightarrow t=-1\frac{dy}{dx}=\frac{1-9t^2}{-6t},{\left(\frac{dy}{dx}\right)}_{t=-1}=\frac{-4}{3}$
Equ of tangent at P is $y-2=\frac{-4}{3}(x+2)$
Q also lies on the above tangent
$\Rightarrow t-3t^3-2=\frac{-4}{3}(1-3t^2+2)\Rightarrow 3t^3+4t^2-t-2=0$
$(3t-2)(t+1{)}^2=0\Rightarrow t=\frac{2}{3}for$ Q
$\therefore Q=(-\frac{1}{3},\frac{-2}{9})$