# Derivation of Hyperbolae

Let the coordinates of a point on a hyperbola be $(x,y)$, the distance from the centre of the hyperbola to either vertex $a$ and the distance from the centre of the hyperbola to either focus $c$.

The difference of the distances from a point on the hyperbola to the two foci is equivalent to both $\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}$ and $2a$, thus $\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=2a$. The equation is true for all points $(x,y)$ on the hyperbola because the difference is invariant of $(x,y)$.

Derivation:

\begin{aligned} \sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}&=2a\\ \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\\ x^2+2cx+c^2+y^2&=4a^2+4a\sqrt{(x-c)^2+y^2}+x^2-2cx+c^2+y^2\\ 2cx&=4a^2+4a\sqrt{(x-c)^2+y^2}-2cx\\ 4cx-4a^2&=4a\sqrt{(x-c)^2+y^2}\\ cx-a^2&=a\sqrt{(x-c)^2+y^2}\\ \left(cx-a^2\right)^2&=a^2\left(x^2-2cx+c^2+y^2\right)\\ c^2x^2-2a^2cx+a^4&=a^2x^2-2a^2cx+a^2c^2+a^2y^2\\ c^2x^2+a^4&=a^2x^2+a^2c^2+a^2y^2\\ c^2x^2-a^2x^2-a^2y^2&=a^2c^2-a^4\\ x^2\left(c^2-a^2\right)-a^2y^2&=a^2\left(c^2-a^2\right)\quad\text{Let }a^2+b^2=c^2\Rightarrow b^2=c^2-a^2\\ x^2b^2-a^2y^2&=a^2b^2\\ \end{aligned}

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

Note by Gandoff Tan
7 months, 2 weeks ago

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