# How do you find the standard form of the equation of the ellipse given the properties foci (0,+-5), vertices (0, +-8)?

Jul 28, 2018

The equation of the ellipse is ${y}^{2} / 64 + {x}^{2} / 39 = 1$

#### Explanation:

The equation of an ellipse with major vertical axis is

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

The center( symmetric wrt the foci and the vertices) of the ellipse is

$C = \left(h , k\right) = \left(0 , 0\right)$

Therefore,

$a = 8$

$c = 5$

${b}^{2} = \left({a}^{2} - {c}^{2}\right) = \left(64 - 25\right) = \sqrt{39}$

The equation of the ellipse is

${y}^{2} / 64 + {x}^{2} / 39 = 1$

graph{(y^2/64+x^2/39-1)=0 [-17.3, 18.75, -8.67, 9.35]}

Jul 28, 2018

The standard equation of vertical ellipse is ${x}^{2} / 39 + {y}^{2} / 64 = 1$

#### Explanation:

The vertices and foci are on the y axis at

$V \left(0 , 8\right) , V ' \left(0 , - 8\right) \mathmr{and} F \left(0 , 5\right) , F ' \left(0 , - 5\right)$

Semi major axis is $a = 8$ and focus $c = 5$ from the center

$\left(0 , 0\right)$. This is vertical ellipsce of which the equation is

${x}^{2} / {b}^{2} + {y}^{2} / {a}^{2} = 1 \mathmr{and} {x}^{2} / {b}^{2} + {y}^{2} / {8}^{2} = 1$

$c = 5$ is the distance from the center to a focus. The relation of

$c , a , b$ is ${c}^{2} = {a}^{2} - {b}^{2} \therefore {5}^{2} = {8}^{2} - {b}^{2}$ or

${b}^{2} = 64 - 25 = 39 \therefore b = \sqrt{39} \approx 6.245$, therefore, the equation

of vertical ellipse is ${x}^{2} / 39 + {y}^{2} / 64 = 1$

graph{x^2/39+y^2/64=1 [-20, 20, -10, 10]} [Ans]