# What is the equation of the parabola that has a vertex at  (-2, 3)  and passes through point  (13, 0) ?

##### 2 Answers
Mar 4, 2018

equation of parabola can be expressed as, $y = a {\left(x - h\right)}^{2} + k$ where, $\left(h , k\right)$ is the coordinate of vertex and $a$ is a constant.

Given,$\left(h , k\right) = \left(- 2 , 3\right)$ and the parabola passes through $\left(13 , 0\right)$,

So,putting the values we get,

$0 = a {\left(13 - \left(- 2\right)\right)}^{2} + 3$

or, $a = - \frac{3}{225}$

So,the equation becomes, $y = - \frac{3}{225} {\left(x + 2\right)}^{2} + 3$ graph{y=(-3/225)(x+2)^2 +3 [-80, 80, -40, 40]}

Mar 4, 2018

$y = - \frac{1}{75} {\left(x + 2\right)}^{2} + 3$

or $x = \frac{5}{3} {\left(y - 3\right)}^{2} - 2$

#### Explanation:

We can make two types of parabolas, one vertical and other horizontal. The equation of vertical parabola, whose vertex is $\left(- 2 , 3\right)$ is

$y = a {\left(x + 2\right)}^{2} + 3$ and as it passes through $\left(13 , 0\right)$, we have

$0 = a {\left(13 + 2\right)}^{2} + 3$ or $a = \frac{- 3}{15} ^ 2 = - \frac{3}{225} = - \frac{1}{75}$

and hence equation is $y = - \frac{1}{75} {\left(x + 2\right)}^{2} + 3$

The curve appears as follows:

graph{(y+1/75(x+2)^2-3)((x+2)^2+(y-3)^2-0.08)=0 [-20, 20, -10, 10]}

The equation of horizontal parabola, whose vertex is $\left(- 2 , 3\right)$ is

$x = a {\left(y - 3\right)}^{2} - 2$ and as it passes through $\left(13 , 0\right)$, we have

$13 = a {\left(0 - 3\right)}^{2} - 2$ or $a = \frac{13 + 2}{3} ^ 2 = \frac{15}{9} = \frac{5}{3}$

and hence equation is $x = \frac{5}{3} {\left(y - 3\right)}^{2} - 2$

The curve appears as follows:

graph{(x-5/3(y-3)^2+2)((x+2)^2+(y-3)^2-0.08)=0 [-20, 20, -10, 10]}