Show that the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ form an ellipse If the definition of an ellipse is the set of points $(x,y)$ such that given two focus points $F_1, F_2$ the sum of the distances from $(x,y)$ to each focus point is constant, how can one show that the curve $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1, \quad 0 < b \leq a \quad$ forms an ellipse?  
The methods that I know of are to either derive the formula by considering the foci $(-c,0), (c,0)$ and the constant distance $2k$, or to set $(x,y) = (a\cos{v}, b\sin{v})$. Is there some other, relatively easy way to show that the points satisfying the equation form an ellipse?  
I am asking because in a book I am reading the author states that a "direct calculation" shows that the curve indeed forms an ellipse, but I do not understand what kind of calculation this might be.  

If $c=\sqrt{a^2-b^2}$ and $F_1=(-c,0)$, $F_2=(c,0)$ are the two foci, then the distance of a generic point $P=(x,y)$ of the ellipse from $F_1$ is
$$
PF_1=\sqrt{(x+c)^2+y^2}=\sqrt{x^2+2cx+c^2+y^2}.
$$
Substituting here $y^2=b^2-{b^2\over a^2}x^2$ one gets
$$
PF_1=\sqrt{x^2+2cx+c^2+b^2-{b^2\over a^2}x^2}=
\sqrt{{c^2\over a^2}x^2+2cx+a^2}
=a+{c\over a}x.
$$
An analogous calculation gives $PF_2=a-{c\over a}x$, so that $PF_1+PF_2=2a$.
