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 $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$.
Best Answer
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 $c^2=a^2-b^2$.