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

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Answer 1 · 2017-01-22T13:45:06

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

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]}

What are the vertices and foci of the ellipse 9x^2-18x+4y^2=279x^2-18x+4y^2=27?