# What is the equation of the parabola that has a vertex at  (-4, 2)  and passes through point  (-12,-3) ?

${\left(y - 2\right)}^{2} = - \frac{25}{8} \left(x + 4\right)$

OR

${\left(x + 4\right)}^{2} = - \frac{64}{5} \left(y - 2\right)$

#### Explanation:

Case 1: Horizontal parabola

Let the equation of horizontal parabola with the vertex $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(- 4 , 2\right)$ be as follows

${\left(y - {y}_{1}\right)}^{2} = 4 a \left(x - {x}_{1}\right)$

${\left(y - 2\right)}^{2} = 4 a \left(x + 4\right)$

Since above parabola passes through the point $\left(- 12 , - 3\right)$ hence it will satisfies above equation as follows

${\left(- 3 - 2\right)}^{2} = 4 a \left(- 12 + 4\right)$

$a = - \frac{25}{32}$

hence, the equation of horizontal parabola

${\left(y - 2\right)}^{2} = 4 \left(- \frac{25}{32}\right) \left(x + 4\right)$

${\left(y - 2\right)}^{2} = - \frac{25}{8} \left(x + 4\right)$

Case 2: Vertical parabola

Let the equation of vertical parabola with the vertex $\left({x}_{1} , {y}_{1}\right) \setminus \equiv \left(- 4 , 2\right)$ be as follows

${\left(x - {x}_{1}\right)}^{2} = 4 a \left(y - {y}_{1}\right)$

${\left(x + 4\right)}^{2} = 4 a \left(y - 2\right)$

Since above parabola passes through the point $\left(- 12 , - 3\right)$ hence it will satisfies above equation as follows

${\left(- 12 + 4\right)}^{2} = 4 a \left(- 3 - 2\right)$

$a = - \frac{16}{5}$

hence, the equation of vertical parabola

${\left(x + 4\right)}^{2} = 4 \left(- \frac{16}{5}\right) \left(y - 2\right)$

${\left(x + 4\right)}^{2} = - \frac{64}{5} \left(y - 2\right)$