Answer:
[tex]\dfrac{x^2}{4}+\dfrac{y^2}{8}=1[/tex]
Step-by-step explanation:
The foci of an ellipse are two distinct points located along the major axis of the ellipse. As the x-coordinates of the given foci (0, 2) and (0, -2) are the same (x = 0), the major axis of the ellipse is vertical.
The foci are equidistant from the center of the ellipse. Therefore:
[tex]\textsf{center}=\left(\dfrac{0+0}{2},\dfrac{2+(-2)}{2}\right) =(0, 0)[/tex]
The general form of a vertical ellipse is:
[tex]\boxed{\begin{array}{l}\underline{\textsf{General equation of a vertical ellipse}}\\\\\dfrac{(x-h)^2}{b^2}+\dfrac{(y-k)^2}{a^2}=1\\\\\textsf{where:}\\\phantom{ww}\bullet \textsf{$2a$ is the major axis.}\\\phantom{ww}\bullet \textsf{$2b$ is the minor axis.}\\\phantom{ww}\bullet \textsf{$(h,k)$ is the center.}\\ \phantom{ww}\bullet \textsf{$(h,k\pm a)$ are the vertices.}\\ \phantom{ww}\bullet \textsf{$(h,k\pm c)$ are the foci where $c^2=a^2-b^2$}\end{array}}[/tex]
As the center of the given ellipse is the origin:
[tex]h = 0[/tex]
[tex]k = 0[/tex]
The formula for the foci of a vertical ellipse is (h, k±c), where c is the distance from the center of the ellipse to each of its foci. Therefore:
[tex](h,k\pm c)=(0,\pm2 ) \\\\(0,0\pm c)=(0,\pm 2)\\\\ (0, \pm c)=(0, \pm 2) \\\\c = 2[/tex]
The eccentricity (e) of an ellipse is the ratio of the distance from the center to a focus (c) to the length of the semi-major axis (a). Therefore:
[tex]e=\dfrac{c}{a} \\\\\\ c=ae[/tex]
As c = 2, then:
[tex]ae = 2[/tex]
The formula for the directrices of a vertical ellipse is:
[tex]y=k\pm \dfrac{a}{e}[/tex]
Given that the directrices are y = ±4 and k = 0:
[tex]0\pm \dfrac{a}{e}=\pm 4 \\\\\\ \dfrac{a}{e}=4 \\\\\\ e=\dfrac{a}{4}[/tex]
Substitute e = a / 4 into ae = 2 and solve for a²:
[tex]a \cdot \dfrac{a}{4}=2 \\\\\\ a^2=8[/tex]
To find the value of b², we can use the formula c² = a² - b²:
[tex]c^2 = a^2 - b^2 \\\\2^2 = 8 - b^2 \\\\b^2 = 8-2^2 \\\\b^2 = 8-4 \\\\b^2 = 4[/tex]
Finally, substitute h = 0, k = 0, a² = 8 and b² = 4 into the general equation of a vertical ellipse:
[tex]\dfrac{(x-0)^2}{4}+\dfrac{(y-0)^2}{8}=1 \\\\\\ \dfrac{x^2}{4}+\dfrac{y^2}{8}=1[/tex]
Therefore, the equation of an ellipse with directrices at y = ±4 and foci at (0, 2) and (0, -2) is:
[tex]\Large\boxed{\boxed{\dfrac{x^2}{4}+\dfrac{y^2}{8}=1}}[/tex]