Question

Let $L$ be a normal to the parabola $y^2=4x$. lf $L$ passes through the point $(9,6)$, then $L$ is given by

• A. $y-x+3=0$
• B. $y+3x-33=0$
• C. $y+x-15=0$
• D. $y-2x+12=0$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $y-x+3=0$
B $y+3x-33=0$
D $y-2x+12=0$

Equation of the normal is $y=mx-2m-m^3$

It passes through the point $(9,6)$ then

$6=9m-2m-m^3$

$\Rightarrow m^3-7m+6=0$

$\Rightarrow (m-1)(m-2)(m+3)=0$

$\Rightarrow m=1,2,-3$

Equations of normals are $y-x+3=0,y+3x-33=0,y-2x+12=0$