Question

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^2-8x+2y+7=0`. What is its eccentricity? Give a rough sketch of this conic.

Answer 1

The vertex is `V=(4,9/2)`
The focus is `F=(4,4)`
The directrix is `y=5`
The eccentricity of a parabola `=1` by definition

Explanation:

Let's rewrite this equation and complete the squares

`x^2-8x+2y+7=0`

`x^2-8x=-2y-7`

`(x^2-8x+16)=-2y-7+16`

`(x-4)^2=-2y+9=-2(y-9/2)`

`(x-4)^2=-2(y-9/2)`

We compare this equation to

`(x-a)^2=2p(y-b)`

The vertex is `V=(a,b)=(4,9/2)`

`p=-1`

The focus is `F=(a,b+p/2)=(4,4)`

The directrix is `y=b-p/2=9/2+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]}