# Question #a3b13

Mar 20, 2017

#### Explanation:

Given: ${y}^{2} - 8 x + 2 y + 25 = 0$

Write in standard form, $x = a {y}^{2} + b y + c$:

$x = \frac{1}{8} {y}^{2} + \frac{2}{8} y + \frac{25}{8}$

$a = \frac{1}{8} , b = \frac{2}{8} , \mathmr{and} c = \frac{25}{8}$

The y coordinate, k, of the vertex and the focus is:

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

$k = \frac{- \frac{2}{8}}{2 \left(\frac{1}{8}\right)}$

$k = - 1$

The x coordinate, h, of the vertex is the equation evaluated at y = k= -1:

$h = \frac{1}{8} {\left(- 1\right)}^{2} + \frac{2}{8} \left(- 1\right) + \frac{25}{8}$

$h = 3$

The vertex is the point $\left(3 , - 1\right)$

Find the focal distance, f:

$f = \frac{1}{4 a}$

$f = \frac{1}{4 \left(\frac{1}{8}\right)}$

$f = 2$

The x coordinate of the focus is the x coordinate of the vertex plus the focal distance and the y coordinate is the same.

The focus is $\left(5 , - 1\right)$

The directrix is a vertical line at the x coordinate of the vertex minus the focal distance:

$x = 3 - 2$

$x = 1$ is the equation of the directrix.

For the x intercept evaluate the equation at $y = 0$:

$x = \frac{25}{8}$

Evaluate the discriminant:

${b}^{2} - 4 \left(a\right) \left(c\right) = {\left(\frac{2}{8}\right)}^{2} - \left(4\right) \left(\frac{1}{8}\right) \left(\frac{25}{8}\right) = \frac{4}{64} - \frac{100}{64} = - \frac{96}{64}$

There are no real y intercepts.