Find the vertices and foci of the ellipse with the given equation: 5x^2+7y^2=35

Answer 1

if you put your equation in standard form

x^2/7+y^2/5=1

your a= sqrt 7 and b=sqrt5
in x^2/a^2+y^2/b^2=1

so see here http://www.mathwarehouse.com/ellipse/focus-of-ellipse.php to work out the focus.
Repost if you get stuck.

Answer 2

divide all terms by 35
x^2/7 + y^2/5 = 1
a = √7, b = √5

vertices: (√7,0) and (-√7,0), (0,√5), (0, -√5)
foci (±c,0) but for this kind of ellipse
c^2 + b^2 = a^2
c^ = 2
c = ±√2
foci : (±√2,0)