Equation of Ellipse in Reduced Form/Cartesian Frame

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Theorem

Let $K$ be an ellipse aligned in a cartesian coordinate 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

EllipseEquation.png

By definition, the foci $F_1$ and $F_2$ of $K$ are located at $\left({-c, 0}\right)$ and $\left({c, 0}\right)$ respectively.

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

By definition, these are located at $\left({-a, 0}\right)$ and $\left({a, 0}\right)$.

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

By definition, these are located at $\left({0, -b}\right)$ and $\left({0, b}\right)$.


Let $P = \left({x, y}\right)$ 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:

\(\displaystyle \sqrt {\paren {x - c}^2 + y^2} + \sqrt {\paren {x + c}^2 + y^2}\) \(=\) \(\displaystyle d = 2 a\) Pythagoras's Theorem
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sqrt {\paren {x + c}^2 + y^2}\) \(=\) \(\displaystyle 2 a - \sqrt {\paren {x - c}^2 + y^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {x + c}^2 + y^2\) \(=\) \(\displaystyle \paren {2 a - \sqrt {\paren {x - c}^2 + y^2} }^2\) squaring both sides
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2 + 2 c x + c^2 + y^2\) \(=\) \(\displaystyle 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2\) expanding
\(\displaystyle \leadsto \ \ \) \(\displaystyle x^2 + 2 c x + c^2 + y^2\) \(=\) \(\displaystyle 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2\) further expanding
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^2 - c x\) \(=\) \(\displaystyle a \sqrt {\paren {x - c}^2 + y^2}\) gathering terms and simplifying
\(\displaystyle \leadsto \ \ \) \(\displaystyle \paren {a^2 - c x}^2\) \(=\) \(\displaystyle a^2 \paren {\paren {x - c}^2 + y^2}^2\) squaring both sides
\(\displaystyle \leadsto \ \ \) \(\displaystyle c^2 x^2 - 2 c x a^2 + a^4\) \(=\) \(\displaystyle a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2\) expanding
\(\displaystyle \leadsto \ \ \) \(\displaystyle c^2 x^2 + a^4\) \(=\) \(\displaystyle a^2 x^2 + a^2 c^2 + a^2 y^2\) simplifying
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^4 - a^2 c^2\) \(=\) \(\displaystyle a^2 x^2 - c^2 x^2 + a^2 y^2\) gathering terms
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^2 \paren {a^2 - c^2}\) \(=\) \(\displaystyle \paren {a^2 - c^2} x^2 + a^2 y^2\) simplifying
\(\displaystyle \leadsto \ \ \) \(\displaystyle a^2 b^2\) \(=\) \(\displaystyle b^2 x^2 + a^2 y^2\) substituting $a^2 - c^2 = b^2$ from $(2)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle 1\) \(=\) \(\displaystyle \frac {x^2} {a^2} + \frac {y^2} {b^2}\) dividing by $a^2 b^2$

$\blacksquare$


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