Question

It $x=t^2$ and $y=2t$ then equation of 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 D $x+y-3=0$

$x=t^2,y=2t$
$\Rightarrow \frac{dx}{dt}=2t,\frac{dy}{dt}=2$
$\Rightarrow \frac{dy}{dx}=\frac{1}{t}$
$\therefore$ Slope of normal is $={\left(\frac{-dx}{dy}\right)}_{t=1}=-t=-1$
Therefore, equation of normal is, $(y-2)=-1(x-1)$
$\Rightarrow x+y=3$
Hence, option 'D' is correct.