Derivation of the Cartesian form for an ellipse

The derivation of the Cartesian form for an ellipse is simple and instructive. One definition of the ellipse is the "locus of all points in a plane whose distances to two fixed points (called the foci) add to the same constant". See ellipse for other definitions.

Let the foci be points $(-c,0)$ and $(c,0)$. Then the ellipse center is $(0,0)$. If $(x,y)$ is any point on the ellipse, and if $d_{1}$ is the distance between $(x,y)$ and $(-c,0)$, and $d_{2}$ is the distance between $(x,y)$ and $(c,0)$ as shown in the figure

then we can define the semi-major axis as

$a\equiv {d_{1}+d_{2} \over 2}.$

The cartesian equation may be derived from this definition. Substituting

$a={{\sqrt {(x-c)^{2}+y^{2}}}+{\sqrt {(x+c)^{2}+y^{2}}} \over 2}.$

To simplify we isolate the radical and square both sides.

${\sqrt {(x+c)^{2}+y^{2}}}=2a-{\sqrt {(x-c)^{2}+y^{2}}}$

$(x+c)^{2}+y^{2}=\left(2a-{\sqrt {(x-c)^{2}+y^{2}}}\right)^{2}$

$(x+c)^{2}+y^{2}=4a^{2}-4a{\sqrt {(x-c)^{2}+y^{2}}}+(x-c)^{2}+y^{2}$

Solving for the root and simplifying:

$(x+c)^{2}+y^{2}-(x-c)^{2}-y^{2}-4a^{2}=-4a{\sqrt {(x-c)^{2}+y^{2}}}$

$-{1 \over 4a}((x+c)^{2}-4a^{2}-(x-c)^{2})={\sqrt {(x-c)^{2}+y^{2}}}$

Swap sides to return to original format and continue:

${\sqrt {(x-c)^{2}+y^{2}}}=-{1 \over 4a}((x+c)^{2}-4a^{2}-(x-c)^{2})$

${\sqrt {(x-c)^{2}+y^{2}}}=-{1 \over 4a}(x^{2}+2xc+c^{2}-4a^{2}-x^{2}+2xc-c^{2})$

${\sqrt {(x-c)^{2}+y^{2}}}=-{1 \over 4a}(4xc-4a^{2})$

${\sqrt {(x-c)^{2}+y^{2}}}=a-{c \over a}x$

A final squaring

$(x-c)^{2}+y^{2}=a^{2}-2xc+{c^{2} \over a^{2}}x^{2}$

$x^{2}-2xc+c^{2}+y^{2}=a^{2}-2xc+{c^{2} \over a^{2}}x^{2}$

$x^{2}+c^{2}+y^{2}=a^{2}+{c^{2} \over a^{2}}x^{2}$

Grouping the x-terms and dividing by $a^{2}-c^{2}\,$

$x^{2}-{c^{2} \over a^{2}}x^{2}+y^{2}=a^{2}-c^{2}$

$x^{2}\left(1-{c^{2} \over a^{2}}\right)+y^{2}=a^{2}-c^{2}$

Where: $1={a^{2} \over a^{2}}$

$x^{2}\left({a^{2}-c^{2} \over a^{2}}\right)+y^{2}=a^{2}-c^{2}$

${x^{2} \over a^{2}}+{y^{2} \over a^{2}-c^{2}}=1$

If x = 0 then

$d_{1}=d_{2}=a={\sqrt {c^{2}+b^{2}}}$

(where b is the semi-minor axis)

Therefore, we can substitute

$b^{2}=a^{2}-c^{2}\,$

And we have our desired equation:

${x^{2} \over a^{2}}+{y^{2} \over b^{2}}=1$

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