The equation of the ellipse that has a center at (0,0) , a focus at (3 , 0) , and a vertex at (5 , 0 ) , is \frac(x^2)(A^2)+\frac(y^2)(B^2)=

The equation is given by :

x ^2 /  25  +  y ^2 /  16

So ...  A = 5 and B = 4

Here's the graph.......https://www.desmos.com/calculator/pfvhokujae

http://www.mathopenref.com/coordgeneralellipse.html

If the centre is (0,0)

Let the end of the horizontal axis is  A(5,0)

The foci is      $$S_1(3,0)$$

If you sketch this it is easy to see that the other foci is      $$S_2(-3,0)$$

Let the point   B(0,b) be the end of the vertical axis.

NOW

$$\\S_1B=BS_2 S_1A+AS_2\;\; must\; \;equal \;\; S_1B+BS_2\\\\ 2+8=S_1B+BS_2\\\\ 10=5+5\\\\ so\\\\ B(0,4)\\\\ $equation of the ellipse$\\\\ \frac{x^2}{5^2}+\frac{y^2}{4^2}=1\\\\ \frac{x^2}{25}+\frac{y^2}{16}=1$$