Question

If $x=t^2\mathrm{and}y=2t$, then equation of the normal at $t=1$ is:

• A. $x+y-3=0$
• B. $x+y-1=0$
• C. $x+y+1=0$
• D. $x+y+3=0$

Answer 1

The correct option is A

$x+y-3=0$

Explanation for the correct option.

Step 1: Find $\frac{dy}{dx}$.

At $t=1$, $x=1\mathrm{and}y=2$.

$\frac{dx}{dt}=2t\mathrm{and}\frac{dy}{dt}=2$

So,

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{dy}{dt}\times \frac{dt}{dx}\\ & =& 2\times \frac{1}{2t}\\ & =& \frac{1}{t}\end{array}$

Step 2: Find the equation of the normal.

Slope of normal is given by $n_m\times \frac{dy}{dx}=-1$

At $t=1$,

$\begin{array}{rcl}{n}_m\times \frac{dy}{dx}& =& -1\\ \Rightarrow n_m& =& -1\end{array}$

Equation of normal is $y-y_1=n_m\left(x-x_1\right)$. Here, $\left(x_1,y_1\right)=\left(1,2\right)$

So, at $t=1$the equation of normal will be,

$\begin{array}{rcl}y-2& =& -1\left(x-1\right)\\ & \Rightarrow & y-2+x-1=0\\ \Rightarrow y+x-3& =& 0\end{array}$

Hence, option A is correct.