Question

If $x=t^2$ 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

$x=t^2$ and y=2t
At t=1, x=1 and y =2
Now $\left(\frac{dy}{dx}\right)=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2}{2t}=\frac{1}{t}\Rightarrow {\left(\frac{dy}{dx}\right)}_{t=1}=1$
$\therefore$ Equation of the normal at (1, 2) is $y-2=-\frac{1}{\frac{dy}{dx}}(x-1)$
$\Rightarrow y-2=-1(x-1)\Rightarrow x+y-3=0$.