# What is the vertex and focus of the parabola described by 2x^2-5x+y+50=0?

Mar 11, 2017

The vertex is $V = \left(\frac{5}{4} , - \frac{375}{8}\right)$
The focus is $F = \left(\frac{5}{4} , - \frac{376}{8}\right)$
The directrix is $y = - \frac{374}{8}$

#### Explanation:

Let's rewrite this equation and complete the squares

$2 {x}^{2} - 5 x + y + 50 = 0$

$2 {x}^{2} - 5 x = - y - 50$

$2 \left({x}^{2} - \frac{5}{2} x\right) = - \left(y + 50\right)$

$\left({x}^{2} - \frac{5}{2} x + \frac{25}{16}\right) = - \frac{1}{2} \left(y + 50\right)$

${\left(x - \frac{5}{4}\right)}^{2} = - \frac{1}{2} \left(y + 50 - \frac{25}{8}\right)$

${\left(x - \frac{5}{4}\right)}^{2} = - \frac{1}{2} \left(y + \frac{425}{8}\right)$

We compare this equation to

${\left(x - a\right)}^{2} = 2 p \left(y - b\right)$

The vertex is $V = \left(a , b\right) = \left(\frac{5}{4} , - \frac{375}{8}\right)$

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

The focus is $F = \left(\frac{5}{4} , b + \frac{p}{2}\right) = \left(\frac{5}{4} , - \frac{376}{8}\right)$

The directrix is $y = b - \frac{p}{2} = - \frac{375}{8} + \frac{1}{8} = - \frac{374}{8}$

graph{(2x^2-5x+y+50)(y+374/8)((x-5/4)^2+(y+375/8)^2-0.001)=0 [-1.04, 7.734, -48.52, -44.13]}