# How do you find the equation of an ellipse with major axis vertical and passes through points (0,4) and (2,0)?

Dec 28, 2016

This problem can only be done, if one assumes that two given points, $\left(0 , 4\right) \mathmr{and} \left(2 , 0\right)$, are the upper vertex on major axis and right vertex on the minor axis, respectively.

#### Explanation:

The general Cartesian equation for an ellipse with a vertical major axis is:

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

Where x and y correspond any point, $\left(x , y\right)$ in the ellipse, h and k correspond to the center point $\left(h , k\right)$, "a" is the length of the semi-major axis, and "b" is the length of the semi-minor axis.

Using the previously stated assumption.

The upper vertex on the major axis will have the coordinates:

$\left(h , k + a\right)$

The right vertex on the minor axis will have the coordinates:

$\left(h + b , k\right)$

Matching these two point to the given points:

$\left(h , k + a\right) = \left(0 , 4\right)$
$\left(h + b , k\right) = \left(2 , 0\right)$

This allows us to write the following equations:

$h = 0 , k + a = 4 , h + b = 2 , \mathmr{and} k = 0$

Substituting 0 for h and k:

$h = 0 , a = 4 , b = 2 , \mathmr{and} k = 0$

Substitute these 4 values into equation [1]:

${\left(y - 0\right)}^{2} / {4}^{2} + {\left(x - 0\right)}^{2} / {2}^{2} = 1 \text{ [2]}$