# How do you find the equation of an ellipse with vertices (0,+-5) and passes through the point (4,2)?

Oct 10, 2017

Because the vertices are vertically oriented, we use the general Cartesian form:

(y-k)^2/a^2+(x-h)^2/b^2=1; a > b" [1]"

where $\left(h , k\right)$ is the center.

#### Explanation:

The general form for vertically oriented vertices are:

$\left(h , k - a\right)$ and $\left(h , k - a\right)$

These general forms and the given vertices $\left(0 , - 5\right) \mathmr{and} \left(0 , 5\right)$ allow us to write 3 equations that can be used to find the values of $h , k , \mathmr{and} a$:

$h = 0$

$k - a = - 5$

$k + a = 5$

$2 k = 0$

$k = 0$

$a = 5$

Substitute these values into equation [1]:

(y-0)^2/5^2+(x-0)^2/b^2=1; a > b" [2]"

Substitute the point $\left(4 , 2\right)$ into equation [2] and solve for b:

(2-0)^2/5^2+(4-0)^2/b^2=1; a > b

$\frac{4}{25} + \frac{16}{b} ^ 2 = 1$

$\frac{16}{b} ^ 2 = \frac{21}{25}$

${b}^{2} = \frac{16 \left(25\right)}{21}$

$b = \frac{20 \sqrt{21}}{21}$

${\left(y - 0\right)}^{2} / {5}^{2} + {\left(x - 0\right)}^{2} / {\left(\frac{20 \sqrt{21}}{21}\right)}^{2} = 1 \text{ [3]}$

Here is a graph of equation [3] and the three points: