Equation of Ellipse in Reduced Form/Cartesian Frame

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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

EllipseEquation.png

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:

$(1): \quad d = 2 a$

Also, from Focus of Ellipse from Major and Minor Axis:

$(2): \quad a^2 - c^2 = b^2$


Then:

\(\ds \sqrt {\paren {x - c}^2 + y^2} + \sqrt {\paren {x + c}^2 + y^2}\) \(=\) \(\ds d\) Pythagoras's Theorem
\(\ds \) \(=\) \(\ds 2 a\) from $(1)$
\(\ds \leadsto \ \ \) \(\ds \sqrt {\paren {x + c}^2 + y^2}\) \(=\) \(\ds 2 a - \sqrt {\paren {x - c}^2 + y^2}\)
\(\ds \leadsto \ \ \) \(\ds \paren {x + c}^2 + y^2\) \(=\) \(\ds \paren {2 a - \sqrt {\paren {x - c}^2 + y^2} }^2\) squaring both sides
\(\ds \leadsto \ \ \) \(\ds x^2 + 2 c x + c^2 + y^2\) \(=\) \(\ds 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + \paren {x - c}^2 + y^2\) Square of Difference
\(\ds \leadsto \ \ \) \(\ds x^2 + 2 c x + c^2 + y^2\) \(=\) \(\ds 4 a^2 - 4 a \sqrt {\paren {x - c}^2 + y^2} + x^2 - 2 c x + c^2 + y^2\) Square of Difference
\(\ds \leadsto \ \ \) \(\ds a^2 - c x\) \(=\) \(\ds a \sqrt {\paren {x - c}^2 + y^2}\) gathering terms and simplifying
\(\ds \leadsto \ \ \) \(\ds \paren {a^2 - c x}^2\) \(=\) \(\ds a^2 \paren {\paren {x - c}^2 + y^2}\) squaring both sides
\(\ds \leadsto \ \ \) \(\ds c^2 x^2 - 2 c x a^2 + a^4\) \(=\) \(\ds a^2 \paren {x^2 - 2 c x + c^2 + y^2}\) Square of Difference
\(\ds \) \(=\) \(\ds a^2 x^2 - 2 c x a^2 + a^2 c^2 + a^2 y^2\) Distributive Laws of Arithmetic
\(\ds \leadsto \ \ \) \(\ds c^2 x^2 + a^4\) \(=\) \(\ds a^2 x^2 + a^2 c^2 + a^2 y^2\) 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
\(\ds \leadsto \ \ \) \(\ds a^2 \paren {a^2 - c^2}\) \(=\) \(\ds \paren {a^2 - c^2} x^2 + a^2 y^2\) Distributive Laws of Arithmetic
\(\ds \leadsto \ \ \) \(\ds a^2 b^2\) \(=\) \(\ds b^2 x^2 + a^2 y^2\) 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

EllipseEquation-2.png


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

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