How do you find the center, foci and vertices of #9x^2+25y^2-36x-50y+61=0#?

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Answer 1 · 2017-01-14T07:03:36

The center is C #=(2,1)#
The foci are F#=(10/3,1)#and F'#=(2/3,1)#
The vertices are A#=(11/3,1)# and A'#=(1/3,1)# and B#=(2,2)# and B'#=(2,0)#

Let 's rearrange the equation by completing the squares

#9x^2+25y^2-36x-50y+61=0#

#9x^2-36x+25y^2-50y=-61#

#9(x^2-4x)+25(y^2-2y)=-61#

#9(x^2-4x+color(red)(4))+25(y^2-2y+color(blue)(1))=-61+color(red)(36)+color(blue)(50)#

#9(x-2)^2+25(y-1)^2=25#

Dividing throughout by #25#

#9(x-2)^2/25+25(y-1)^2/25=25/25#

#(x-2)^2/(5/3)^2+(y-1)^2/1=1#

This is the equation of an ellipse

#(x-h)^2/a^2+(y-k)^2/b^2=1#

The center is C #(h,k)=(2,1)#

#c=sqrt(a^2-b^2)=sqrt(25/9-1)=sqrt(16/9)=4/3#

The foci are F#(h+c,k)=(2+4/3,1)=(10/3,1)#

and F'#(h-c,k)=(2-4/3,1)=(2/3,1)#

graph{(x-2)^2/(25/9)+(y-1)^2=1 [-5.546, 5.55, -2.773, 2.774]}