Find the coordinates of the vertices and foci, eccentricity? x28x+2y+7=0

Find the coordinates of the vertices and foci, and the equations of the directrices and axes of the conic x28x+2y+7=0. What is its eccentricity? Give a rough sketch of this conic.

1 Answer

The vertex is V=(4,92)
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

x28x+2y+7=0

x28x=2y7

(x28x+16)=2y7+16

(x4)2=2y+9=2(y92)

(x4)2=2(y92)

We compare this equation to

(xa)2=2p(yb)

The vertex is V=(a,b)=(4,92)

p=1

The focus is F=(a,b+p2)=(4,4)

The directrix is y=bp2=92+12=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]}