How to prove the parametric equation of an ellipse?

Let $OA=a$ and $OB=b$ be the radii of the two circles, and let $C$, $C'$ be the foci of the ellipse, where $OC=OC'=c=\sqrt{a^2-b^2}$. If $H$ is the projection of $A$ on major axis $DE$ and $P$ is the projection of $B$ on $AH$, then you want to show that $PC+PC'=2a$.

Suppose, without loss of generality, that $C$ is the focus nearest to $P$. We have $PC^2=PH^2+CH^2$, but $PH=(b/a)AH$, $HC=|c-OH|$ and $AH^2+OH^2=a^2$, so that: $$ \begin{aligned} PC^2&={b^2\over a^2}AH^2+(c-OH)^2={b^2\over a^2}(a^2-OH^2)+c^2+OH^2-2cOH\\ &=a^2+{c^2\over a^2}OH^2-2cOH=\left(a-{c\over a}OH\right)^2,\\ \end{aligned} $$ and then $PC=a-(c/a)OH$. An analogous computation yields $PC'=a+(c/a)OH$, so that $PC+PC'=2a$, QED.

Why not simply

\begin{equation*} \left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=\cos ^{2}t+\sin ^{2}t=1 \end{equation*}