# How do you write an equation of a ellipse with vertices (0,2), (4,2), and endpoints of the minor axis (2,3), (2,1)?

Nov 29, 2017

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

#### Explanation:

Here we have an ellipse with vertices at $\left(0 , 2\right) \mathmr{and} \left(4 , 2\right)$ and the minor axis between $\left(2 , 3\right) \mathmr{and} \left(2 , 1\right)$

The equation of an ellipse centred at the origin with major axis length $\left(2 a\right)$ and minor axis length $\left(2 b\right)$ is:

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$

Here we have an ellipse with major axis on $y = 2$
(Since the vertices are both on $y = 2$)

And minor axis on $x = 2$
(Since the endpoints of the minor axis are both on $x = 2$)

Hence, the centre of the ellipse is at $\left(2 , 2\right)$

From the above we can calculate the length of the major axis to be:
$4 - 0 = 4$

$\to a = \frac{4}{2} = 2$

and the length of the minor axis to be:
$3 - 1 = 2$

$\to b = \frac{2}{2} = 1$

Applying the shifts of the centre from $\left(0 , 0\right)$ to $\left(2 , 2\right)$ our standard equation becomes:

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

Substituting for $a \mathmr{and} b$ above

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

Hence, the equation of our ellipse is: ${\left(x - 2\right)}^{2} / 4 + {\left(y - 2\right)}^{2} / 1 = 1$

We can see this ellipse in the graphic below that satisfies the given conditions.

graph{(x-2)^2/4+(y-2)^2/1=1 [-3.26, 7.84, -0.51, 5.04]}