$\begingroup$I have summarized the question below:
If the vertices of an ellipse centered at the origin are $(a,0),(-a,0),(0,b),$ and $(0,-b)$, and $a>b$, prove that for foci at $(\pm c,0)$, $c^2=a^2-b^2$.
I am guessing that I have to use the distance formula, which is $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
$\endgroup$12 Answers
$\begingroup$With drawing ellipse shape with center at the origin and are $A(\pm a,0)$ and $B(0,\pm b)$ are vertices, find a symmetric shape and symmetric foci at $F(c,0)$ and $F'(-c,0)$. With definition for ellipse for exery point $X$ on ellipse $|XF|+|XF'|=2a$ so $|BF|+|BF'|=2a$ so $2\sqrt{b^2+c^2}=2a$ that concludes $c^2=a^2-b^2$.
$\endgroup$3$\begingroup$Any point on the ellipse is such that $MF_1+MF_2=AF_1+AF_2=2a$ where $F_1,F_2$ are the foci and $A$ is the $(a,0)$ vertex. So let's write that for $B(0,b)$
$$\sqrt{c^2+b^2}+\sqrt{c^2+b^2}=2a$$
This rewrites easily as $c^2+b^2=a^2$. QED
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