What are the vertices and foci of the ellipse #9x^2-18x+4y^2=27#?

1 Answer
Jan 22, 2017

Answer:

The vertices are #(3,0), (-1,0), (1,3), (1,-3)#
The foci are #(1,sqrt5)# and #(1,-sqrt5)#

Explanation:

Let's rearrange the equation by completing the squares

#9x^2-18x+4y^2=27#

#9(x^2-2x+1)+4y^2=27+9#

#9(x-1)^2+4y^2=36#

Dividing by #36#

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

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

This is the equation of an ellipse with a vertical major axis

Comparing this equation to

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

The center is #=(h,k)=(1,0)#

The vertices are A#=(h+a,k)=(3,0)# ; A'#=(h-a,k)=(-1,0)# ;

B#=(h.k+b)=(1,3)# ; B'#=(h,k-b)=(1,-3)#

To calculate the foci, we need

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

The foci are F#=(h.k+c)=(1, sqrt5)# and F'#=(h,k-c)=(1,-sqrt5)#

graph{(9x^2-18x+4y^2-27)=0 [-7.025, 7.02, -3.51, 3.51]}