Ellipse/Proofs

The derivation of the cartesian form for an ellipse is simple and instructive. An ellipse is defined as a the loci of points equidistant to two fixed points called the foci. Assuming that the foci are located at (-c,0) and (c,0) (ie. the ellipse is centered at (0,0)) then the sum of the distance between any point (x,y) and the two foci is constant. and you always need to remember the general equation for horizontal and vertical ellipses.

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), i.e.

then we can define $a$

:$d\_1\; +\; d\_2\; =\; 2a,$

(a here is the semi-major axis, although this is irrelevant for the sake of the proof). From this simple definition we can derive the cartesian equation. Substituting:

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

To simplify we isolate the radical and square both sides.

:$sqrt\; \{(x+c)^2+y^2\}\; =\; 2a\; -\; sqrt\; \{(x-c)^2+y^2\}$:$(x+c)^2\; +\; y^2\; =\; left\; (\; 2a\; -\; sqrt\{(x-c)^2+y^2\}\; ight\; )^2$:$(x+c)^2\; +\; y^2\; =\; 4a^2\; -\; 4asqrt\{(x-c)^2+y^2\}\; +\; (x-c)^2\; +y^2$

Solving for the root and simplifying:

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

A final squaring

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

Grouping the x-terms and dividing by $a^2-c^2$

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

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:

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