Ellipse with Horizontal Major Axis

Major axis along the x-axis, length 2a. Minor axis along the y-axis, length 2b. Vertices at (±a, 0) and (0, ±b). Foci at (±c, 0) where c² = a² − b².

Let the two foci be $S(-c, 0)$ and $S'(c, 0)$. The ellipse is the locus of $P(x, y)$ such that $PS + PS' = 2a$ (constant, with $2a > 2c$, i.e. $a > c$).

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

Isolate one radical and square:

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

Squaring again:

$$a^4 - 2a^2cx + c^2x^2 = a^2[(x-c)^2 + y^2]$$ $$a^4 - 2a^2cx + c^2x^2 = a^2x^2 - 2a^2cx + a^2c^2 + a^2y^2$$ $$a^4 - a^2c^2 = a^2x^2 - c^2x^2 + a^2y^2$$ $$a^2(a^2-c^2) = (a^2-c^2)x^2 + a^2y^2$$

Setting $b^2 = a^2 - c^2 > 0$ (since $a > c$):

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

Elements: Vertices $(±a, 0)$, co-vertices $(0, ±b)$, foci $(±c, 0)$, $c = \sqrt{a^2-b^2}$.