Question

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

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

Answer 1

Answer: (A), (B), (C)

$y–2x+12=0$

The equation of the normal to the given parabola $y^2=4x$ in slope form is
$y=mx–2m–m^3$
which will pass through $(9,6)$ if
$6=9m–2m–m^3$
$\Rightarrow m^3–7m+6=0$
$\Rightarrow m=1,2,–3$
Consequently the equation of the normal $L$ is
$y=x–3\Rightarrow y–x+3=0$
or $y=2x–12\Rightarrow y–2x+12=0$
or $y=–3x+33\Rightarrow y+3x–33=0$