Hyperbola with Horizontal Transverse Axis

Transverse axis along the x-axis, length 2a. Conjugate axis along the y-axis, length 2b. Vertices at (±a, 0). Foci at (±c, 0) where c² = a² + b².

A hyperbola is the locus of all points $P(x, y)$ such that $|PS - PS'| = 2a$ (constant), where $S$ and $S'$ are the two foci and $2a < 2c = SS'$.

Setup: Let $S(-c, 0)$ and $S'(c, 0)$. For the right branch, $PS' - PS = 2a$ (closer to $S'$):

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

Isolate one radical and square (same algebraic process as the ellipse derivation, with $2a < 2c$):

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

For this to be valid, $cx + a^2 < 0$, which holds on the right branch when $x > 0$ and $c > a$. Squaring:

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

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

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

Summary of elements: