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

Jan 22, 2017

The vertices are $\left(3 , 0\right) , \left(- 1 , 0\right) , \left(1 , 3\right) , \left(1 , - 3\right)$
The foci are $\left(1 , \sqrt{5}\right)$ and $\left(1 , - \sqrt{5}\right)$

Explanation:

Let's rearrange the equation by completing the squares

$9 {x}^{2} - 18 x + 4 {y}^{2} = 27$

$9 \left({x}^{2} - 2 x + 1\right) + 4 {y}^{2} = 27 + 9$

$9 {\left(x - 1\right)}^{2} + 4 {y}^{2} = 36$

Dividing by $36$

${\left(x - 1\right)}^{2} / 4 + {y}^{2} / 9 = 1$

${\left(x - 1\right)}^{2} / {2}^{2} + {y}^{2} / {3}^{2} = 1$

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

Comparing this equation to

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$

The center is $= \left(h , k\right) = \left(1 , 0\right)$

The vertices are A$= \left(h + a , k\right) = \left(3 , 0\right)$ ; A'$= \left(h - a , k\right) = \left(- 1 , 0\right)$ ;

B$= \left(h . k + b\right) = \left(1 , 3\right)$ ; B'$= \left(h , k - b\right) = \left(1 , - 3\right)$

To calculate the foci, we need

$c = \sqrt{{b}^{2} - {a}^{2}} = \sqrt{9 - 4} = \sqrt{5}$

The foci are F$= \left(h . k + c\right) = \left(1 , \sqrt{5}\right)$ and F'$= \left(h , k - c\right) = \left(1 , - \sqrt{5}\right)$

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