Focus of Ellipse from Major and Minor Axis

Theorem

Let $K$ be an ellipse whose major axis is $2 a$ and whose minor axis is $2 b$.

Let $c$ be the distance of the foci of $K$ from the center.

Then:

$a^2 = b^2 + c^2$

Proof

Let the foci of $K$ be $F_1$ and $F_2$.

Let the vertices of $K$ be $V_1$ and $V_2$.

Let the covertices of $K$ be $C_1$ and $C_2$.

Let $P = \tuple {x, y}$ be an arbitrary point on the locus of $K$.

From the equidistance property of $K$ we have that:

$F_1 P + F_2 P = d$

where $d$ is a constant for this particular ellipse.

This is true for all points on $K$.

In particular, it holds true for $V_2$, for example.

Thus:

\(\ds d\)

\(=\)

\(\ds F_1 V_2 + F_2 V_2\)

\(\ds \)

\(=\)

\(\ds \paren {a + c} + \paren {a - c}\)

\(\ds \)

\(=\)

\(\ds 2 a\)

It also holds true for $C_2$:

$F_1 C_2 + F_2 C_2 = d$

Then:

\(\ds F_1 C_2^2\)

\(=\)

\(\ds O F_1^2 + O C_2^2\)

Pythagoras's Theorem

\(\ds \)

\(=\)

\(\ds c^2 + b^2\)

and:

\(\ds F_1 C_2^2\)

\(=\)

\(\ds O F_1^2 + O C_2^2\)

Pythagoras's Theorem

\(\ds \)

\(=\)

\(\ds c^2 + b^2\)

Thus:

\(\ds F_1 C_2 + F_2 C_2\)

\(=\)

\(\ds 2 \sqrt {b^2 + c^2}\)

\(\ds \)

\(=\)

\(\ds 2 a\)

as $2 a = d$

\(\ds \leadsto \ \ \)

\(\ds a\)

\(=\)

\(\ds \sqrt {b^2 + c^2}\)

\(\ds \leadsto \ \ \)

\(\ds b^2 + c^2\)

\(=\)

\(\ds a^2\)

$\blacksquare$

Also presented as

This result is also seen presented as:

$c^2 = a^2 - b^2$