Standard Equation of an Ellipse

Deriving the standard equation $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ (with $a>b$).

Let $S$ be the focus, $d$ the directrix, and $e$ the eccentricity ($0<e<1$) of the ellipse. Draw $SZ$ perpendicular to the directrix. Let $A$ and $A'$ divide the segment $SZ$ internally and externally, respectively, in the ratio $e:1$ — by the very definition of the ellipse, both $A$ and $A'$ satisfy $SA=e\cdot AZ$ and $SA'=e\cdot A'Z$, so both points lie on the ellipse.

Let $AA'=2a$, and take the midpoint $O$ of $AA'$ as the origin, with $OS$ as the $X$-axis. Then $O\equiv(0,0)$, $A\equiv(a,0)$, $A'\equiv(-a,0)$.

Writing $Z\equiv(k,0)$ and $S\equiv(h,0)$, the internal/external section-formula relations

lead (after solving these two equations simultaneously) to $k=\dfrac{a}{e}$ and $h=ae$. So the focus is $S(ae,0)$ and the foot of the perpendicular is $Z(a/e,0)$; the directrix is $x=\dfrac{a}{e}$.

Now let $P(x,y)$ be any point on the ellipse. Computing $SP$ and the perpendicular distance $PM$ to the directrix:

Substituting into $SP=e\cdot PM$, then squaring both sides and simplifying:

Since $0<e<1$, we have $1-e^2>0$, so we may safely divide both sides by $a^2(1-e^2)$:

Writing $b^2=a^2(1-e^2)$ (with $a>b$, since $1-e^2<1$), this becomes the standard equation of the ellipse:

$\boxed{\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1}, \qquad b^2=a^2(1-e^2),\ a>b.$ …