Question

The straight line which is both tangent and normal to the curve $x=3t^2,y=2t^3$ is

• A. $y+\sqrt{3}(x-1)=0$
• B. $y-\sqrt{3}(x-1)=0$
• C. $y+\sqrt{2}(x-2)=0$
• D. $y-\sqrt{2}(x-2)=0$

Answer 1

Answer: (C), (D)

The correct options are
C $y+\sqrt{2}(x-2)=0$
D $y-\sqrt{2}(x-2)=0$

$x=3t^2$ and $y=2t^3$
$\Rightarrow \frac{dy}{dx}=t$

Let the line be $y=mx+c$
$\Rightarrow 2t^3=3m{t}^2+c$
$\Rightarrow t^3-\frac{3}{2}m{t}^2-\frac{c}{2}=0$ ... (1)

The slope of the line is $m.$ Therefore, the slope of the line at the repeated root is also $m$ and this is the point $t=m.$ The slope of the curve at the other point is $-\frac{1}{m}$(because it is normal to the line) and therefore the other solution is $t=-\frac{1}{m}$.

So the cubic expression must be identically equal to:
$(t-m{)}^2(t+\frac{1}{m})=(t^2-2mt+m^2)(t+\frac{1}{m})$
$=t^3+(\frac{1}{m}-2m)t^2+(m^2-2)t+m$ ... (2)

(1) and (2) are same equations so we can compare coefficients.
$\Rightarrow -\frac{3}{2}m=\frac{1}{m}-2m$
$\Rightarrow 0=m^2-2$ and $-\frac{1}{2}c=m\Rightarrow c=-2m$
$\Rightarrow m=\pm \sqrt{2}$

Hence, equation of the line is:
$y=\pm \sqrt{2}(x-2)$