Find the standard form of the equation of the ellipse with the given characteristics and center the origin.
vertices= (0, + -8)
foci= (0, + -4)
by standard form you may mean the Canonical implicit equation which is
$\displaystyle \frac{\left(x-h\right)^2}{a^2} + \frac{\left(y-k\right)^2}{b^2} = 1$
Foci: $\displaystyle \left(h \pm c, k\right)$
Vertices:$\displaystyle \left(h \pm a, k)\right)$
where $\displaystyle c^2= a^2- b^2$
just plug away
the standard or general equation actually is
$\displaystyle A X^2 + B X Y + C Y^2 + D X + E Y + F = 0$
You will also need to know that $\displaystyle c^2= a^2- b^2$.
To see that, remember that an ellipse can be defined as the locus of points whose total distance from the two foci is constant. Take the point at one end of the major axis (the axis containing the foci). The distance from one focus to the center is c, then the distance from the center to the vertex is a. The distance back from the that vertex to the other focus is a- c. The total distance is a+ c+ (a- c)= 2a.
Now, look at the vertex at one end of axis perpendicular to the axis containing the foci. The line from each focus to that vertex is the hypotenuse of the right triangle formed by that vertex, the center, and the focus. It has legs of length b and c and so the length of the hypotenuse is $\displaystyle \sqrt{b^2+ c^2}$. Since that is true for both foci, the total distance from the foci to that vertex is $\displaystyle 2\sqrt{b^2+ c^2}$. Since that total distance is constant, we must have $\displaystyle 2\sqrt{b^2+ c^2}= 2a$. Divide both sides by 2 and square both sides to get $\displaystyle b^2+ c^2= a^2$.
just plug away
the standard equation actually is
$\displaystyle A X^2 + B X Y + C Y^2 + D X + E Y + F = 0$