Answer
$\frac{x^2}{4} + \frac{4y^2}{7}=1$
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
$a \gt b$
2a = length of major axis
2b = length of minor axis
c = distance of each focus from the center
$a^2=b^2+c^2$
The foci are (-1.5, 0) and (1.5, 0), which are equidistant from the origin (0, 0). This means that the ellipse has its center at (0, 0).
Each focus is 1.5 units away from the center therefore $c=1.5$.
RECALL:
The Eccentricity (e) of an ellipse is given by the formula:
$\\e=\frac{c}{a}$
Substitute the known values of $e$ and $c$ to have:
$\\e= \frac{c}{a}
\\0.75 = \frac{1.5}{a}
\\a(0.75) = 1.5
\\a=\frac{1.5}{0.75}
\\a=2$
Solve for $b$ using the formula $a^2=b^2+c^2$ to have:
$a^2=b^2+c^2
\\2^2=b^2+(1.5)^2
\\4=b^2+2.25
\\4-2.25=b^2
\\1.75=b^2
\\\sqrt{1.75}=b
\\\sqrt{\frac{7}{4}}=b
\\\frac{\sqrt7}{2}=b$
When connected together using a line, the foci form a horizontal line. Thus, the major axis is also horizontal.
Therefore, the equation of the ellipse is:
$\\\frac{x^2}{a^2} + \frac{y^2}{b^2}=1
\\\frac{x^2}{2^2}+\frac{y^2}{(\frac{\sqrt7}{2})^2}=1
\\\frac{x^2}{4}+\frac{y^2}{\frac{7}{4}}=1
\\\frac{x^2}{4} + y^2 \cdot \frac{4}{7}=1
\\\frac{x^2}{4} + \frac{4y^2}{7}=1$
