# How do you find the center and vertices of the ellipse 4x^2+y^2=1?

Dec 14, 2016

The center is $= \left(0 , 0\right)$
The vertices are $\left(\frac{1}{2} , 0\right)$,$\left(- \frac{1}{2} , 0\right)$,$\left(0 , 1\right)$ and $\left(0 , - 1\right)$

#### Explanation:

Let's rewrite the equation

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

The equation of the ellipse, center, $\left(h , k\right)$

is

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

So, the center is $\left(0 , 0\right)$

To find the vertices, let's $y = 0$

Then, $x = \pm \frac{1}{2}$

The vertices are $\left(\frac{1}{2} , 0\right)$ and $\left(- \frac{1}{2} , 0\right)$

Let $x = 0$, $\implies$, $y = \pm 1$

The other vertices are $\left(0 , 1\right)$ and $\left(0 , - 1\right)$
graph{4x^2+y^2=1 [-2.16, 2.165, -1.082, 1.08]}