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Q.
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
NTA AbhyasNTA Abhyas 2020Conic Sections
Solution:
$4x-4y-21=0$
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 $\left(x - 3\right)=1\left(y + 3\right)-\frac{2}{4}\left(1\right)-\frac{1}{4}\left(1\right)^{3}$
$\Rightarrow 4x-4y-21=0$