# How do you solve the standard form of an ellipse given Foci: (0, -3) , (0, 3) Vertices: (0, -4), (0, 4)?

Oct 24, 2016

${x}^{2} / 7 + {y}^{2} / 16 = 1$

#### Explanation:

${f}_{1} : \left(0 , - 3\right)$

${f}_{2} : \left(0 , 3\right)$

${V}_{1} : \left(0 , - 4\right)$

${V}_{2} : \left(0 , 4\right)$

Distance ${D}_{f}$ between foci $\implies 2 c$
Distance ${D}_{V}$ between vertices $\implies 2 a$

$D = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

${D}_{f} = \sqrt{{\left(0 - 0\right)}^{2} + {\left(- 3 - 3\right)}^{2}}$
$\implies {D}_{f} = \sqrt{- {6}^{2}} = 6$
$\implies c = 3$

${D}_{V} = \sqrt{{\left(0 - 0\right)}^{2} + {\left(- 4 - 4\right)}^{2}}$
$\implies {D}_{V} = \sqrt{- {8}^{2}} = 8$
$\implies a = 4$

${c}^{2} = {a}^{2} - {b}^{2}$
$\implies {b}^{2} = {a}^{2} - {c}^{2}$

${b}^{2} = {4}^{2} - {3}^{2}$
${b}^{2} = 16 - 9 = 7$

$\implies b = \sqrt{7}$

Midpoint $M$ between the foci and/or vertices $\implies$ center $C : \left(h , k\right)$ of the ellipse

$M : \left(\frac{{x}_{1} + {x}_{2}}{2} , \frac{{y}_{1} + {y}_{2}}{2}\right)$

Using either the foci or the vertices, we get

$C : \left(0 , 0\right)$

Note that between the foci and between the vertices, only the y-coordinate changes. This means the ellipse has a vertical major axis

Standard Equation of an ellipse with vertical major axis

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

${\left(x - 0\right)}^{2} / {\left(\sqrt{7}\right)}^{2} + {\left(y - 0\right)}^{2} / \left({4}^{2}\right) = 1$

$\implies {x}^{2} / 7 + {y}^{2} / 16 = 1$