Let $L$ be a normal to the parabola $y^2=4 x$. If $L$ passes through the point $(9,6)$, then $L$ is given by
Let $L$ be a normal to the parabola $y^2=4 x$. If $L$ passes through the point $(9,6)$, then $L$ is given by
$y-x+3=0$
$y+3 x-33=0$
$y+x-15=0$
$y-2 x+12=0$
Solution
Normal to $y^2=4 x$, is $y-m x-2 m-m^3$ which passes through $(9,6)$.
Now, $\quad 6=9 m-2 m-m^3$ $\Rightarrow m^3-7 m+6=0 \Rightarrow m=1,2,-3$
$\therefore$ Equation of normals are
$y-x+3=0$
$y+3 x-33=0$ and $y-2 x+12=0$