Question

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

• A. $3x-y=6$
• B. $3x-y=27$
• C. $3x-y=33$
• D. $3x-y=21$

Answer 1

The correct option is C $3x-y=33$

$\because$ Required normal is perpendicular to the line $x+3y+1=0$
Hence slope of the normal is $m=3$
Any normal to the given parabola is
$y=mx-2am-a{m}^3$
Here $a=1$ and $m=3$
$\therefore$ Required normal equation is
$y=3x-33\Rightarrow 3x-y=33$