Answer
$\frac{x^2}{4} + \frac{y^2}{3}=1$
Work Step by Step
RECALL:
The standard equation of an ellipse with whose center is at (0, 0) is:
(i) $\frac{x^2}{a^2} + \frac{y^2}{b^2}=1$ (horizontal major axis)
(ii) $\frac{x^2}{b^2} + \frac{y^2}{a^2}=1$ (vertical major axis)
where
$a \gt b$
a = distance of each vertex from the center
b = distance of the each minor axis endpoint from the center
c= distance of each focus from the center
$a^2=b^2+c^2$
The given ellipse has its vertices at $(-2, 0)$ and $(2, 0)$. Since these points are equidistant (2 units away) from the origin (0, 0), then it means that the center of the ellipse is at the origin.
Note that:
(1) Each vertex is 2 units away from the center so $a=2$.
(2) Each focus is 1 unit away from the center so $c=1$.
(3) The vertices are horizontal so the major axis is horizontal.
Solve for $b$ by using the formula $a^2=b^2+c^2$ to have:
$2^2 = b^2+1^2
\\4=b^2+1
\\4-1=b^2
\\3=b^2
\\\sqrt3=b$
Therefore the equation of the ellipse is:
$\frac{x^2}{2^2} + \frac{y^2}{(\sqrt3)^2} = 1
\\\frac{x^2}{4} + \frac{y^2}{3}=1$
