# How do you find an equation of the parabola with a vertex of (-1,2) and focus of (-1,0)?

Feb 13, 2015

The answer is: $y = - \frac{1}{8} {x}^{2} - \frac{1}{4} x + \frac{15}{8}$

First of all some formulas:

$y = a {x}^{2} + b x + c$ is the equation of a parabola with axis of symmetry parallel to the y-axis;

$V \left(- \frac{b}{2 a} , - \frac{\Delta}{4 a}\right)$ where $\Delta = {b}^{2} - 4 a c$, is the vertex;

$F \left(- \frac{b}{2 a} , \frac{1 - \Delta}{4 a}\right)$ is the focus;

$x = - \frac{b}{2 a}$ is the axis of symmetry;

$y = \frac{- 1 - \Delta}{4 a}$ is the directrix.

So we have to solve this system of equations:

$- \frac{b}{2 a} = - 1$

$- \frac{\Delta}{4 a} = 2$

$\frac{1 - \Delta}{4 a} = 0$

Than:

$b = 2 a$

$\Delta = - 8 a$

$\Delta = 1$

And:

$a = - \frac{1}{8}$

$b = - \frac{1}{4}$

$\Delta = 1$

From the last one we have to find the value of $c$

${b}^{2} - 4 a c = 1 \Rightarrow \frac{1}{16} - 4 \cdot \left(- \frac{1}{8}\right) \cdot c = 1 \Rightarrow \frac{1}{16} + \frac{1}{2} c = 1 \Rightarrow$

$\frac{1}{2} c = 1 - \frac{1}{16} \Rightarrow c = 2 \cdot \frac{15}{16} \Rightarrow c = \frac{15}{8}$.

The equation becomes: $y = - \frac{1}{8} {x}^{2} - \frac{1}{4} x + \frac{15}{8}$