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.

1 Answer

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]}