# How do you write an equation of an ellipse in standard form given Center at (-3,1); Vertex at (-3,3); Focus at (-3,0)?

Dec 29, 2015

#### Answer:

${\left(x + 3\right)}^{2} / 4 + {\left(y - 1\right)}^{2} / 3 = 1$

#### Explanation:

When the major axis is horizontal, the standard equation of an ellipse is

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

where
$C : \left(h , k\right)$
$V : \left(h \pm a , k\right)$

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

$f : \left(h \pm c , k\right)$

On the other hand, when the major axis is vertical, the standard equation of an ellipse is

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

where
$C : \left(h , k\right)$
$V : \left(h , k \pm a\right)$

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

$f : \left(h , k \pm c\right)$

In the given,
$C : \left(- 3 , 1\right)$
${V}_{1} : \left(- 3 , 3\right)$
${f}_{1} : \left(- 3 , 0\right)$

Since the $x$-coordinate of the points are constant, we can say that the major axis is vertical.

$C : \left(h , k\right)$
$C : \left(- 3 , 1\right)$

$\implies h = - 3$
$\implies k = 1$

$V : \left(h , k \pm a\right)$

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

$\implies 3 = k \pm a$
$\implies 3 = 1 \pm a$

$\implies 3 = 1 + a$
$\implies 3 = 1 - a$

$\implies 2 = a$
$\implies - 2 = a$

Since $a$ is the distance between the center and the vertex, we take $a$ to be positive

$\implies a = 2$

$f : \left(h , k \pm c\right)$

$\implies 0 = 1 \pm c$
$\implies c = 1$
$\implies c = - 1$

Similarly for $c$, since it is the distance between the focus and the center, we take $c$ to be positive

$\implies c = 1$

${c}^{2} = {a}^{2} - {b}^{2}$
${1}^{2} = {2}^{2} - {b}^{2}$

$\implies 1 = 4 - {b}^{2}$
$\implies {b}^{2} = 3$
$\implies b = \pm \sqrt{3}$

Again, since $b$ is the distance between the center and the end of the minor axis (not sure what it was called), we take $b$ to be positive

$\implies b = \sqrt{3}$

Hence, the equation of the ellipse is

${\left(x - - 3\right)}^{2} / {2}^{2} + {\left(y - 1\right)}^{2} / {\left(\sqrt{3}\right)}^{2} = 1$

$\implies {\left(x + 3\right)}^{2} / 4 + {\left(y - 1\right)}^{2} / 3 = 1$