Question

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

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

Answer 1

Answer: (A), (B)

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

The slope of the tangent at $P\left(t\right)=\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=t$
Let the tangent cuts the curve and normal at $Q\left(t_1\right)$, then the coordinate of the point is $(3t_1^2,2t_1^3)$
The slope of the line passing through $P$ and $Q$=$\frac{(2t_1^3-2t^3)}{3t_1^2-3t^2}=t$
$\Rightarrow t_1=-\frac{t}{2}$and $Q=(3\frac{t^2}{4},-\frac{t^3}{4})$
Slope of normal at $Q=-\frac{dx}{dy}=\frac{2}{t}$
The equation of straight line having slope$t$ and passing through$Q$ is $y-(-\frac{t^3}{4})=t(x-3\frac{t^2}{4})$
$\Rightarrow tx-y=t^3\cdots \cdots \left(1\right)$
The equation of line having slope $\frac{2}{t}$and passing through $Q$ is $y-(-\frac{t^3}{4})=\frac{2}{t}(x-3\frac{t^2}{4})$
$\Rightarrow \frac{2x}{t}-y=\frac{3t}{2}+\frac{t^3}{4}\cdots \cdots \left(2\right)$
On solving $\left(1\right)$ and $\left(2\right)$we have $t=\sqrt{2}$
$\Rightarrow$ Option $\left(A\right)$ and $\left(B\right)$ are correct