Question

The equation of a normal to the parabola $y=x^2-6x+6$ which is perpendicular to the line joining the origin to the vertex of the parabola is

• A. $4x-4y-11=0$
• B. $4x-4y+1=0$
• C. $4x-4y-21=0$
• D. $4x-4y+21=0$

Answer 1

Answer: (C)

$4x-4y-21=0$
Here $y={\left(x-3\right)}^2-3$
$\Rightarrow y+3={\left(x-3\right)}^2$
Hence, the vertex is $\left(3,-3\right)$
Slope of the line joining origin $\left(0,0\right)$ and vertex $\left(3,-3\right)$ is $-1$
Hence, the slope of the normal is $1$
Equation of the normal is $\left(x-3\right)=1\left(y+3\right)-\frac{2}{4}\left(1\right)-\frac{1}{4}{\left(1\right)}^3$
$\Rightarrow 4x-4y-21=0$