# Find the coordinates of the vertices and foci, eccentricity? x^2-8x+2y+7=0

## Find the coordinates of the vertices and foci, and the equations of the directrices and axes of the conic ${x}^{2} - 8 x + 2 y + 7 = 0$. What is its eccentricity? Give a rough sketch of this conic.

##### 1 Answer
Mar 11, 2017

The vertex is $V = \left(4 , \frac{9}{2}\right)$
The focus is $F = \left(4 , 4\right)$
The directrix is $y = 5$
The eccentricity of a parabola $= 1$ by definition

#### Explanation:

Let's rewrite this equation and complete the squares

${x}^{2} - 8 x + 2 y + 7 = 0$

${x}^{2} - 8 x = - 2 y - 7$

$\left({x}^{2} - 8 x + 16\right) = - 2 y - 7 + 16$

${\left(x - 4\right)}^{2} = - 2 y + 9 = - 2 \left(y - \frac{9}{2}\right)$

${\left(x - 4\right)}^{2} = - 2 \left(y - \frac{9}{2}\right)$

We compare this equation to

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

The vertex is $V = \left(a , b\right) = \left(4 , \frac{9}{2}\right)$

$p = - 1$

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

The directrix is $y = b - \frac{p}{2} = \frac{9}{2} + \frac{1}{2} = 5$

The eccentricity of a parabola $= 1$ by definition

graph{(x^2-8x+2y+7)(y-5)((x-4)^2+(y-4)^2-0.01)=0 [-16.02, 16.01, -8.01, 8.01]}