Question

The equation of the normal to the parabola $y^2=4x$ which is perpendicular to $x+3y+1=0$ is

• A. $3x-y=33$
• B. $3x-y+33=0$
• C. $3x+y=33$
• D. $3x+y+33=0$

Answer 1

Answer: (A)

Given equation of parabola is $y^2=4x$.
The equation of the normal to the parabola
$y^2=4ax$ at $\left(a{m}^2,-2am\right)$ is
$y=mx-2am-a{m}^3$
where $m$ is the slope of the normal.
Here, $a=1$, so the equation of the normal is
$y=mx-2m-m^3$
Since, the normal is perpendicular to the line
$x+3y+1=0$
Then, $m=$ slope of normal to line $x+3y+1=0$ is
$m\times \left(-\frac{1}{3}\right)=-1$
$\Rightarrow m=3$
$\therefore y=3x-2(3)-(3{)}^3$
$\Rightarrow y=3x-6-27$
$\Rightarrow y=3x-33$
$\Rightarrow 3x-y=33$