# How do you find center, vertex, and foci of an ellipse x^2 + 4y^2 = 1?

Nov 15, 2015

$C : \left(0 , 0\right)$
$V : \left(\pm 1 , 0\right)$
$f : \left(\pm \frac{\sqrt{3}}{2} , 0\right)$

#### Explanation:

The standard equation of an ellipse is either in the form

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

or

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

where $a > b$

In the given equation

${x}^{2} + 4 {y}^{2} = 1$

This is equivalent to

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

Our center is at $\left(h , k\right)$

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

Since $a$ is under $x$, the major axis is horizontal. The vertex is at

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

${V}_{1} : \left(0 + 1 , 0\right) \implies \left(1 , 0\right)$

${V}_{2} : \left(0 - 1 , 0\right) \implies \left(- 1 , 0\right)$

Meanwhile, the foci are $c$ units from the center.
Where ${c}^{2} = {a}^{2} - {b}^{2}$

$\implies {c}^{2} = {1}^{2} - {\left(\frac{1}{2}\right)}^{2}$

$\implies {c}^{2} = 1 - \frac{1}{4}$

$\implies {c}^{2} = \frac{4 - 1}{4}$

$\implies {c}^{2} = \frac{3}{4}$

$\implies c = \frac{\sqrt{3}}{2}$

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

${f}_{1} : \left(0 + \frac{\sqrt{3}}{2} , 0\right) \implies \left(\frac{\sqrt{3}}{2} , 0\right)$

${f}_{2} : \left(0 - \frac{\sqrt{3}}{2} , 0\right) \implies \left(- \frac{\sqrt{3}}{2} , 0\right)$