Conic Ellipse problem For an ellipse, $9x^2+4y^2=36$, find vertices and foci. 
I would first standardised the equation in form $(x)^2/a^2 + y^2/b^2=1$ thus…………(i)
divide all sides by 36,I get:
$$(9x^2)/36+(4y^2)/36=36/36$$ which is equiv. to $x^2/4+y^2/9=36/36$
based on  $x^2/4+y^2/9=36/36$  we can rewrite this as:
$$x^2/2^2 +y^2/3^2 =1$$   which is in standard  $x^2/a^2 +y^2/b^2 =1$
we can even pluck out our values for a,b thus: a=2 and b=3
but by pythagoras,we know $c^2=a^2-b^2$ hence $c^2=2^2-3^2$
so in my case, $c^2=4-9= -5$ and $\therefore$  $c=√(-5)$
But I reckon the values of a,b can swap depending on  major axis since we can re-write standard equation as $$(x)^2/b^2 + y^2/b^2=1$$ which will bring a different value of $c^2$. 
Questions: How do I make sure I get the correct value of $c^2$?
Question: How do ensure the ellipse is plotted on the correct major axis?


For the ellipse $9x^2 + 4y^2 = 36$ ...

This is, as you have found, equivalent to $$\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$$
Now in an ellipse, the larger denominator corresponds with $a$.
(Contrast this with a hyperbola, where the denominator of the positive variable term is $a^2$)
So we find that $a = 3, b = 2$. The major axis is vertical, as the $a$ was under the $y^2$.
Therefore, the vertices are up and down $(a = 3)$ units from the center, which is the origin in this case.
The covertices are $(b = 2)$ units left and right of center. 
$c = \sqrt5$, and the foci are $c$ units up and down from the center. 
