Remember, finding the vertices is quite simple.[br][br]You simply add and subtract the major value from it's respected coordinate of the center.[br]Horizontal: [math]\left(h+a,k\right)[/math] and [math]\left(h-a,k\right)[/math][br]Vertical: [math]\left(h,k+b\right)[/math] and [math]\left(h,k-b\right)[/math][br][br]To find the foci it's quite similar.[br]Horizontal: [math]\left(h+c,k\right)[/math] and [math]\left(h-c,k\right)[/math][br]Vertical: [math]\left(h,k+c\right)[/math] and [math]\left(h,k-c\right)[/math][br][br]The next question you may have is: how do I find c?[br][math]c^2=L^2-S^2[/math], this means, larger square minus the smaller square.[br][br]As an example, take the ellipse: [math]\frac{\left(x-2\right)^2}{9}+\frac{\left(y+3\right)^2}{25}=1[/math].[br]Note that the center is (2,-3) and the vertices are (2,2) and (2,-8).[br][br]To find the foci, the larger square is 25, and the smaller is 9. [br]so [math]c^2=25-9[/math][br][math]c^2=16[/math][br][math]c=4[/math][br][br]So the foci are (2,-3[math]\pm[/math]4) which is (2,1) and (2,-7)