Find the equation of the parabola given the tangent to a point and another point.

Question

I have a problem with derivatives, I've been trying to solve but I was not able to do it.

A parabola is tangent to the line $3x-y+6 = 0$ in the point $(0,6)$ and goes through the point $(1,0)$. Find the equation of the parabola supposing the equation is of the form $y = Ax^2 +Bx +C$ where $A B C$ are unknown.

I would like you to help me to solve it. So far I know $(1,0)$ is the vertex of the parabola and that $(0,6)$ is a point of the parabola,thus the directive is $x=0$

Any help would be appreciated.

Answer 1

From the tangent line, we see that the parabola has y'(0)=3 at (0,6). This gives: $$y'(0)=2A(0)+B=3$$$$B=3$$ From y(0)=6, we get: $$y(0)=A(0)^2+B(0)+C=6$$ $$C=6$$ From y(1)=0, we get: $$y(1)=A(1)^2+3(1)+6=0$$ $$A+9=0$$ $$A=-9$$

Answer 2

If you already have a tangent line, then taking the derivative of your given parabola form will yield

$$y'=2Ax+B$$

and it should be easy to determine the values of $A$ and $B$. But you are also given a nice point on the parabola, namely $(1,0)$, and this should tell you what $C$ is.