# How do you write the equation of the parabola in vertex form given vertex (5,-2) and focus (5,-4)?

Jan 22, 2018

Equation is $y = - \frac{1}{8} {\left(x - 5\right)}^{2} - 2$

#### Explanation:

As vertex is $\left(5 , - 2\right)$ and focus is $\left(5 , - 4\right)$

as axis of symmetry passes through both and abscissa is common in both, equation of axis of symmetry is $x = 5$

and as directrix is perpendicular to it, its equation is of type $y = k$

Now vertex is midway between directrix and focus and hence directrix is $y = 0$

The parabola is locus of a point $\left(x , y\right)$ which moves so that it is equidistant from directrix anf focus, hence its equation is

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

or ${\left(x - 5\right)}^{2} + {y}^{2} + 8 y + 16 = {y}^{2}$

or $8 y = - {\left(x - 5\right)}^{2} - 16$

i.e. $y = - \frac{1}{8} {\left(x - 5\right)}^{2} - 2$

graph{y(x^2-10x+8y+41)(x-5)((x-5)^2+(y+2)^2-0.03)((x-5)^2+(y+4)^2-0.03)=0 [-5.04, 14.96, -7.16, 2.84]}