Equation of Ellipse in Reduced Form/Cartesian Frame
Theorem
Let $K$ be an ellipse aligned in a cartesian plane in reduced form.
Let:
- the major axis of $K$ have length $2 a$
- the minor axis of $K$ have length $2 b$.
The equation of $K$ is:
- $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$
Proof 1
By definition, the foci $F_1$ and $F_2$ of $K$ are located at $\tuple {-c, 0}$ and $\tuple {c, 0}$ respectively.
Let the vertices of $K$ be $V_1$ and $V_2$.
By definition, these are located at $\tuple {-a, 0}$ and $\tuple {a, 0}$.
Let the covertices of $K$ be $C_1$ and $C_2$.
By definition, these are located at $\tuple {0, -b}$ and $\tuple {0, b}$.
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.
From Equidistance of Ellipse equals Major Axis:
- $d = 2 a$
Also, from Focus of Ellipse from Major and Minor Axis:
- $a^2 - c^2 = b^2$
Then:
\(\ds \sqrt {\paren {x - c}^2 + y^2} + \sqrt {\paren {x + c}^2 + y^2}\) | \(=\) | \(\ds d = 2 a\) | Pythagoras's Theorem | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sqrt {\paren {x + c}^2 + y^2}\) | \(=\) | \(\ds 2 a - \sqrt {\paren {x - c}^2 + y^2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {x + c}^2 + y^2\) | \(=\) | \(\ds \paren {2 a - \sqrt {\paren {x - c}^2 + y^2} }^2\) | squaring both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2 + 2 c x + c^2 + y^2\) | \(=\) | \(\ds 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2\) | expanding | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2 + 2 c x + c^2 + y^2\) | \(=\) | \(\ds 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2\) | further expanding | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^2 - c x\) | \(=\) | \(\ds a \sqrt {\paren {x - c}^2 + y^2}\) | gathering terms and simplifying | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {a^2 - c x}^2\) | \(=\) | \(\ds a^2 \paren {\paren {x - c}^2 + y^2}^2\) | squaring both sides | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds c^2 x^2 - 2 c x a^2 + a^4\) | \(=\) | \(\ds a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2\) | expanding | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds c^2 x^2 + a^4\) | \(=\) | \(\ds a^2 x^2 + a^2 c^2 + a^2 y^2\) | simplifying | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^4 - a^2 c^2\) | \(=\) | \(\ds a^2 x^2 - c^2 x^2 + a^2 y^2\) | gathering terms | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^2 \paren {a^2 - c^2}\) | \(=\) | \(\ds \paren {a^2 - c^2} x^2 + a^2 y^2\) | simplifying | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds a^2 b^2\) | \(=\) | \(\ds b^2 x^2 + a^2 y^2\) | substituting $a^2 - c^2 = b^2$ from $(2)$ | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds 1\) | \(=\) | \(\ds \frac {x^2} {a^2} + \frac {y^2} {b^2}\) | dividing by $a^2 b^2$ |
$\blacksquare$
Proof 2
Let $P$ be an arbitrary point in the plane.
Let $PM$ be dropped perpendicular to $V_1 V_2$.
Hence $M = \tuple {x, 0}$.
From Intersecting Chord Theorem for Conic Sections:
- $PM^2 = k V_1 M \times M V_2$
for some constant $k$.
Hence:
\(\ds y^2\) | \(=\) | \(\ds k \paren {a + x} \paren {a - x}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds k \paren {a^2 - x^2}\) |
Putting $x = 0$, we find the points where $K$ crosses the $y$-axis are defined by:
- $y^2 = k a^2$
This gives us:
- $k a^2 = b^2$
and so:
\(\ds y^2\) | \(=\) | \(\ds \dfrac {b^2} {a^2} \paren {a^2 - x^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds b^2 - x^2 \dfrac {b^2} {a^2}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \dfrac {y^2} {b^2}\) | \(=\) | \(\ds 1 - \dfrac {x^2} {a^2}\) |
Hence the result.
$\blacksquare$
Sources
- Weisstein, Eric W. "Ellipse." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/Ellipse.html