Answer:
\(\dfrac{x^2}{144}+\dfrac{y^2}{80}=1\)
Step-by-step explanation:
The foci of an ellipse lie on its major axis, which is the longest diameter of the ellipse. As the y-coordinates of the foci (-8, 0) and (8, 0) are the same (y = 0), this means that the major axis is parallel to the x-axis, and the ellipse is horizontal.
The general equation for a horizontal ellipse is:
\(\boxed{\begin{array}{l}\underline{\textsf{General equation of a horizontal ellipse}}\\\\\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^2}{b^2}=1\\\\\textsf{where:}\\\phantom{ww}\bullet \textsf{$2a$ is the major axis.}\\\phantom{ww}\bullet \textsf{$2b$ is the minor axis.}\\\phantom{ww}\bullet \textsf{$(h,k)$ is the center.}\\ \phantom{ww}\bullet \textsf{$(h\pm a,k)$ are the vertices.}\\ \phantom{ww}\bullet \textsf{$(h\pm c, k)$ are the foci where $c^2=a^2-b^2$}\end{array}}\)
Given that the center (h, k) is at the origin (0, 0), then:
\(h = 0\\\\k = 0\)
Substitute the values of h = 0 and k = 0 into the vertices formula to find the value of a:
\((h\pm a, k) =(0\pm 12, 0)\\\\(0 \pm a, 0)=(0\pm 12, 0)\\\\(\pm a, 0)=(\pm 12, 0)\\\\a=12\)
Therefore, the value of a² is:
\(a^2=12^2\\\\a^2=144\)
Substitute the values of h = 0 and k = 0 into the foci formula to find the value of c:
\((h\pm c, k) =(0\pm 8, 0)\\\\(0 \pm c, 0)=(0\pm 8, 0)\\\\(\pm c, 0)=(\pm 8, 0)\\\\c=8\)
To find the value of b², substitute the values of a and c into the formula c² = a² - b²:
\(c^2=a^2-b^2\\\\8^2=12^2-b^2\\\\64=144-b^2\\\\b^2=144-64\\\\b^2=80\)
Finally, substitute the values of h, k, a² and b² into the general equation of a vertical ellipse:
\(\dfrac{(x-0)^2}{144}+\dfrac{(y-0)^2}{80}=1\\\\\\\dfrac{x^2}{144}+\dfrac{y^2}{80}=1\)
Therefore, the equation of the ellipse centered at the origin with vertices located at (-12, 0) and (12 ,0) and foci at (-8, 0) and (8, 0) is:
\(\large\boxed{\boxed{\dfrac{x^2}{144}+\dfrac{y^2}{80}=1}}\)
