getting the equation of ellipse using the unit circle. Some days ago ,I wondered how to find the equation of ellipse using circle.
so this was my idea.
the equation of a the unit circle with center $(0,0)$ is $x^2+y^2=1$.
So $y=- \sqrt{1-x^2}$ or $y=+ \sqrt{1-x^2}$.
So I considered the difference between a circle and ellipse is that  the ellipse have not a fixed distance between it center and other points.so my idea is starting with unit circle and make the radius not fix.exactly my idea is  compressing or stretching(or both,not with same ratio )the graph of $f(x)=- \sqrt{1-x^2}$ and $f(x)=+ \sqrt{1-x^2}$,because that will give me a shape which is  looks like ellipse .
we know that the graph of $g$ with $g(x)=f(\frac{1}{a}x)$ is just the same graph of $f$ but it compressed to the $y$ axis (if $\frac{1}{a}>1$)  or stretched (if $\frac{1}{a}<1$) so if we want to compressing or stretching the graph of $f(x)= \sqrt{1-x^2}$ we can do this instead $f(x)= \sqrt{1-(\frac{1}{a}x)^2}$.(the same for the other sign)
we know that the graph of $g$ with $g(x)=bf(x)$ is just the same graph of $f$ but it compressed to the $x$ axis (if $b>1$)  or stretched (if $b<1$) so if we want to compressing or stretching the graph of $f(x)= \sqrt{1-(\frac{1}{a}x)^2}$ we can do this instead $f(x)= b\sqrt{1-(\frac{1}{a}x)^2}$.(the same for the other sign)
wo we will have $y^2=b^2(1-(\frac{1}{a}x)^2)$ with some manipulation we will have this $\frac{x^2}{a^2} +\frac{y^2}{b^2}=1$.(now with this steps we can say $2a$ is the Highest width of ellipse and $2b$ is the  Highest height of our ellipse  )
so we will have this equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ wich looks like the exact equation of an ellipse with center $(0,0)$.(if we want the equation of ellipses with other center we can just apply the idea behind $g(x)=f(x+c)$ and $g(x)=f(x)+c$).
so as it seem ,i get the right equation ,but does the idea and steps is correct ,because i did not talk about the two  "focus", i did not even considered  the definition of ellipse ,my idea is  just considered  the ellipse as a distorted circle.
 A: First of all, nice work deriving the standard equation for the ellipse by stretching the circle independently in each coordinate axis.

To answer your question about the foci of the ellipse: there are several definitions of an ellipse, and one can prove that they are all equivalent.
One of those definitions is the one you're working with: a circle of radius $1$ whose radius is stretched by independent factors $a$ and $b$ in perpendicular directions yielding the familiar equation
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. 
$$
Another definition is the locus of points $P$ with the property that the sum of the distances $|F_1P| + |F_2P|$ is constant, where $F_1$ and $F_2$ are the foci.
Let's assume $a \geq b$, otherwise we have to swap the roles of $x$ and $y$. It turns out that if you place $F_1$ and $F_2$ on the $x$-axis, a distance $c$ from the origin, where
$$
c^2 = a^2 - b^2, 
\tag{1}
$$
and you use $2a$ for total distance, i.e.
$$
|F_1P| + |F_2P| = 2a, 
\tag{2}
$$
then the equation of the ellipse again works out to be
$$
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. 
$$

It's not obvious a priori that these two definitions define the same curve in the plane. But, happily, they do! It just takes equations $(1)$ and $(2)$, the Euclidean distance formula, and some algebra to prove it. Give it a try!
