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

Question

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.

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Answer 1 · 2017-03-11T08:50:14

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

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]}