Derive the equation of the normal to the parabola $y =x^2-6x+6$ perpendicular to the straight line joining the origin to the vertex of the parabola.
Let us find the slope of the parabola by differentiating both sides of equation $ \frac{dy}{dx} = 2x-6$
Let the vertex of the parabola be P $(x_1 ,y_1)$ slope of the line joining point P and origin is $\frac{y_2-y_1}{x_2-x_1}$ here $ x_2 = y_2 =0$ being origin.
therefore slope is $\frac{y_1}{x_1}$
This slope is perpendicular to the given normal ( whose equation is required)
Please guide further.. Thanks..
Best Answer
In general, a parabola can be written in the form $$(y - a)= k(x - b)^2$$
where $(a, b)$ is the vertex of the parabola, and $k$ is some non-zero constant.
$$y =x^2-6x+6 \iff (y - 3) = x^2 - 6x + 9 \iff (y + 3) = (x - 3)^2$$
Hence, you have a parabola whose vertex is $(3, -3)$, which opens up.
Can you use what you know now?
Also, if the slope of a tangent line is $m$, then the slope of the line normal to the tangent line is $\frac{-1}{m}$.