How do you find the center, foci and vertices of #9x^2+4y^2-36x+8y+31=0#?

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Answer 1 · 2018-06-17T07:01:06

Please see the explanation below

The equation is

#9x^2+4y^2-36x+8y+31=0#

#<=>#, #9x^2-36x+4y^2+8y+31=0#

#<=>#, #9(x^2-4x)+4(y^2+2y)=-31#

Complete the squares

#<=>#, #9(x^2-4x+4)+4(y^2+2y+1)=-31+36+4#

#<=>#, #9(x-2)^2+4(y+1)^2=9#

Dividing by #9#

#<=>#, #(x-2)^2+(y+1)^2/(9/4)=1#

This is the equation of an ellipse, center #(2,-1)#

The vertices are #A=(2,0.5)#, #A'=(2,-2.5)#, #B=(3,-1)# and #B'=(1,-1)#

The semi axes are #a=1# and #b=3/2#

#c=sqrt(b^2-a^2)=sqrt(9/4-1)=sqrt5/2#

The foci are #F=(2,sqrt5/2)# and #F'=(2,-sqrt5/2)#

graph{9x^2+4y^2-36x+8y+31=0 [-3.05, 5.718, -3.592, 0.79]}