# Reduced equation of the horizontal hyperbola

Reduced equation of the hyperbola. This set is formed by the hyperbolas which symmetry axes correspond with the coordinated axes, and that therefore it also sees its center coinciding with the coordinated origin.

In the first term, we deal with the reduced horizontal hyperbolas, which are those in which the abscissa axis corresponds with the focal axis. The foci will be then at points $$F'(-c,0)$$ and $$F(c,0)$$.

Applying these foci to the general definition of the hyperbola $$\overline{PF}-\overline{PF'}=2a$$$we obtain the expression $$\displaystyle \sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=2a$$$

On having added the root, and squaring it: $$\begin{array}{rcl} \Big(\sqrt{(x+c)^2+y^2}\Big)^2 & = & \Big(2a+\sqrt{(x-c)^2+y^2}\Big)^2 \\ (x+c)^2+y^2 & = & 4a^2+4a\sqrt{(x-c)^2+y^2}+(x-c)^2+y^2 \\ x^2+2xc+c^2+y^2 & = & 4a^2+4a \sqrt{(x-c)^2+y^2}+x^2-2xc+c^2+y^2 \end{array}$$$Simplifying and dividing by four: $$\begin{array}{rcl} 4xc & = & 4a^2+4a \sqrt{(x-c)^2+y^2} \\ xc & = & a^2+a \sqrt{(x-c)^2+y^2} \end{array}$$$

On having cleared the root and having squared it again: $$\displaystyle \begin{array}{rcl} (cx-a^2)^2 & = & (a\sqrt{(x-c)^2+y^2})^2 \\ c^2x^2-2a^2cx+a^4& = & a^2((x-c)^2+y^2) \\ c^2x^2-2a^2cx+a^4 & = & a^2(x^2-2cx+c^2+y^2) \\ c^2x^2-2a^2cx+a^4 & = & a^2x^2-2a^2cx+a^2c^2+a^2y^2) \\ c^2x^2-a^2x^2-a^2y^2& = & a^2c^2-a^4 \\ (c^2-a^2)x^2-a^2y^2 & = & a^2(c^2-a^2)\end{array}$$$We then divide by $$a^2(c^2-a^2)$$ to get $$1$$ on the right: $$\displaystyle \begin{array}{rcl} \frac{(c^2-a^2)x^2}{a^2(c^2-a^2)}-\frac{a^2y^2}{a^2(c^2-a^2)} & = & 1 \\ \frac{x^2}{a^2} - \frac{y^2}{(c^2-a^2)}& = & 1 \end{array}$$$

Applying the definition $$c^2=a^2+b^2$$, $$b^2=c^2-a^2$$ is substituted and we get the desired equation: $$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$$$Next, an example to show the development within a practical example: Find the equation of the reduced horizontal hyperbola with foci in $$F' (-2,0)$$ and $$F (2,0)$$ and eccentricity $$2$$. With the foci, we identify $$c=2$$. As $$\displaystyle e=\frac{c}{a}$$ , then $$a=1$$. On having applied, it $$\overline{PF}-\overline{PF'}=2$$ we arrive at $$\sqrt{(x+2)^2 + y^2}-\sqrt{(x+2)^2 + y^2}=2$$$

The steps of the demonstration then follow: $$\displaystyle \begin{array} {rcl} (\sqrt{(x+2)^2 + y^2})^2 & = & (2+\sqrt{(x+2)^2 + y^2})^2 \\ (x+2)^2 + y^2 & = & 2\cdot 2+2\cdot 2\sqrt{(x+2)^2 + y^2}+(x+2)^2 + y^2 \\ x^2+2x\cdot +2^2+y^2 & = & 4+4\sqrt{(x+2)^2 + y^2}+x^2-2x \cdot 2+2^2+y^2 \end{array}$$$It is simplified and divided by $$4$$: $$\begin{array}{rcl} 8x & = & 4+4\sqrt{(x-2)^2+y^2} \\2x & = & 1+\sqrt{(x-2)^2+y^2} \\ 2x-1 & = & \sqrt{(x-2)^2+y^2}\end{array}$$$

We raise it again to the square to undo the root.$$\begin{array}{rcl} (2x-1)^2 & = & (\sqrt{(x-2)^2+y^2})^2 \\ 2^2x^2-2 \cdot 2 \cdot x +1^2 & = & (x-2)^2+y^2 \\ 4x^2-4x+1 & = & x^2-2 \cdot 2 x+2^2+y^2 \\ 4x^2-4x+1 & = & x^2-4x+4+y^2 \\ 3x^2-y^2 & = & 3\end{array}$$$and finally, dividing by $$3$$, we get the equation of the hyperbola: $$\displaystyle x^2-\frac{y^2}{3}=1$$$

Consider the equation $$\displaystyle \frac{x^2}{16}-\frac{y^2}{9}=1$$\$, and find:

a) The focal distance

To identify in $$\displaystyle \frac{x^2}{a^2}-\frac{y^2}{b^2}=1, a^2=16$$ and $$b^2= 9$$.

With $$c^2=a^2+b^2, c^2=16+9=25$$ and $$c=5$$.

The focal distance is $$2c=10$$.

b) The position of the apexes

$$a= \sqrt{16}=4$$

The apexes are in $$A' (-a, 0)$$ and $$A (a, 0)$$. Identifying $$A' (-4,0)$$ and $$A (4,0)$$.

c) The eccentricity

The eccentricity is $$\displaystyle e=\frac{c}{a}=\frac{5}{4}$$.