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

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]}