Show that the curve $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ form an ellipse

Question

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.

Answer 1

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 $PF_1+PF_2=2a$.

Answer 2

You can write a point on the curve as $(a\cos t,b\sin t)$ for a unique $t\in[0,2\pi)$. Consider the distanc from $(c,0)$, where $c=\sqrt{a^2-b^2}$: \begin{align} \sqrt{(a\cos t-c)^2+(b\sin t)^2} &=\sqrt{a^2\cos^2t-2ac\cos t+a^2-b^2+b^2\sin^2t} \\ &=\sqrt{a^2\cos^2t-2ac\cos t+a^2-b^2+b^2\sin^2t} \\ &=\sqrt{c^2\cos^2t-2ac\cos t+a^2} \\ &=|c\cos t-a| \\ &=a-c\cos t \end{align} Similarly, the distance from $(-c,0)$ is $$ a+c\cos t $$ so the sum is $2a$.