Focal Properties and Latus Rectum of an Ellipse

This section collects the key derived facts about the standard ellipse, each following from the defining relations $b^2=a^2(1-e^2)$ and the two-focus, two-directrix picture of section 7.2.1.

1. Distance between the directrices. Since the two directrices are $x=a/e$ and $x=-a/e$, the distance between them is simply

2. End points of the latus rectum. Let $LSL'$ be the latus rectum through the (right) focus $S(ae,0)$; write $L\equiv(ae,l)$ for the upper end point. Since $L$ lies on the ellipse:

Since $b^2=a^2(1-e^2)$, we have $1-e^2=b^2/a^2$, so $l^2=b^2\cdot\dfrac{b^2}{a^2}=\dfrac{b^4}{a^2}$, giving $l=\dfrac{b^2}{a}$ (taking the positive value, first quadrant). So $L\equiv\left(ae,\dfrac{b^2}{a}\right)$ and $L'\equiv\left(ae,-\dfrac{b^2}{a}\right)$.

3. Length of the latus rectum. By symmetry, $l(LSL')=l(SL)+l(SL')=\dfrac{b^2}{a}+\dfrac{b^2}{a}=\dfrac{2b^2}{a}$.

4. Sum of focal distances is constant — the "pin and string" property. For a point $P$ on the ellipse, let $SP$ and $S'P$ be its distances to the two foci, and $PM,PM'$ its distances to the two directrices. By the focus–directrix property applied to each focus: $SP=e\cdot PM$ and $S'P=e\cdot PM'$. Adding:

So the sum of the focal distances of any point on the ellipse is the constant $2a$ — the length of the major axis. This is precisely the classical "pin-and-string" construction: fix two pins at $S_1,S_2$ with $d(S_1S_2)<K$, loop a string of total length $K=2a$ around them, and trace a pencil kept taut against the string — the resulting curve is an ellipse with foci $S_1,S_2$.

5. Auxiliary circle. The circle drawn with the major axis $AA'$ as its diameter (so centred at the origin, radius $a$) is called the auxiliary circle of the ellipse.

6. Parametric form and the eccentric angle. Let $P(x,y)$ be a point on the ellipse, and let $Q$ be the point on the auxiliary circle directly above/below $P$ (i.e. $PQ\perp$ major axis, sharing the same $x$-coordinate). Writing $\angle XOQ=\theta$, the point $Q$ on the circle of radius $a$ is $Q=(a\cos\theta,\,a\sin\theta)$, so $P=(a\cos\theta,\,y)$ for some $y$ to be found. Substituting into the ellipse equation:

So $P(x,y)\equiv(a\cos\theta,\,b\sin\theta)\equiv P(\theta)$. This is the parametric form $x=a\cos\theta,\,y=b\sin\theta$, and $\theta$ is called the eccentric angle of $P$ (note: $\theta$ is generally not the actual angle that $OP$ makes with the $X$-axis — that would need $\tan(\angle POX)=y/x=(b/a)\tan\theta$, a different angle unless $a=b$). Solving for $\theta$ in terms of a given $(x,y)$: $\theta=\tan^{-1}\left(\dfrac{ay}{bx}\right)$.

7. The vertical ellipse (b > a). $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $b>a$ is the OTHER standard form — here the major axis is along $Y$ instead of $X$. Every formula above still applies, but with the roles of $a,b$ and $X,Y$ interchanged (see the comparison table on this section).

Worked Example 1 — full property set for four given ellipses.

1. $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$: $a^2=16,b^2=9\Rightarrow a=4,b=3$; $e^2=1-\dfrac9{16}=\dfrac7{16}\Rightarrow e=\dfrac{\sqrt7}{4}$; foci $\left(\pm\sqrt7,0\right)$; directrices $x=\pm\dfrac{16}{\sqrt7}$; latus rectum $=\dfrac{2(9)}{4}=\dfrac92$; parametric form $x=4\cos\theta,y=3\sin\theta$.
2. $3x^2+4y^2=1$, i.e. $\dfrac{x^2}{1/3}+\dfrac{y^2}{1/4}=1$: $a^2=\dfrac13,b^2=\dfrac14\Rightarrow a=\dfrac1{\sqrt3},b=\dfrac12$ ($a>b$); $e^2=1-\dfrac{b^2}{a^2}=1-\dfrac34=\dfrac14\Rightarrow e=\dfrac12$; centre $(0,0)$; vertices $\left(\pm\dfrac1{\sqrt3},0\right)$ and $\left(0,\pm\dfrac12\right)$; foci $\left(\pm\dfrac1{2\sqrt3},0\right)$.
3. $4x^2+3y^2=1$, i.e. $\dfrac{x^2}{1/4}+\dfrac{y^2}{1/3}=1$: here $a^2=\dfrac14,b^2=\dfrac13$ with $b>a$, so the $Y$-axis is the major axis; $e^2=1-\dfrac{a^2}{b^2}=1-\dfrac34=\dfrac14\Rightarrow e=\dfrac12$; foci on the $Y$-axis at $(0,\pm be)$.
4. $4x^2+9y^2-16x+54y+61=0$: completing the square gives $4(x-2)^2+9(y+3)^2=36$, i.e. $\dfrac{(x-2)^2}{9}+\dfrac{(y+3)^2}{4}=1$: $a^2=9,b^2=4\Rightarrow a=3,b=2$; centre $(2,-3)$; $e^2=1-\dfrac49=\dfrac59\Rightarrow e=\dfrac{\sqrt5}{3}$; vertices at $(2\pm3,-3)=(5,-3),(-1,-3)$ and $(2,-3\pm2)=(2,-1),(2,-5)$. …