**Hyperbola** is not a closed curve, but its parts are in a way bounded by two lines –* asymptotes*. Do you know that some astronomical objects and specific protons are moving in a way we can’t explain using only circles and ellipses? For these movements it is important to introduce another* second degree curve*.

It is uniquely defined by its **foci** and a **vertex**.

Suppose points $F_1$ and $ F_2$ are solid points of a plain $M$ and a a positive real number such that $ a < \frac{1}{2} \mid F_1 F_2 \mid$. Hyperbola is a set of all points of a plane $M$ for which the absolute value of the differences from points $F_1$ and $ F_2$ is a constant and equal to $2a$, where a is the distance of a vertex to the center of a hyperbola.

This definition may seem a little bit complicated so we’ll explain it using a drawing.

Let’s choose any point $B$.

$\mid F_2 B \mid – \mid B F_1 \mid = 2 \cdot 3 = 6$

This equality will be valid for all points on a hyperbola. In other words, we can define it as follows:

$\{ T \in \mathbb{M} : \mid \mid T F_1 \mid – \mid T F_2 \mid \mid = 2a \}$

## How to draw a hyperbola?

First, you need to find the ** foci** $ (F_1, F_2)$,

**(point $O$) and**

*centre***($A$, $B$).**

*vertexes*** Centre of the hyperbola** is the middle point of a line ${F_1 F_2}$.

Line ${AB}$ is called the ** real axis** and lines ${OA}$ and ${OB}$ are called

*.*

**real semi-axis**Number $ e = \frac{1}{2} \mid F_1 F_2 \mid$ is called * linear eccentricity of a hyperbola*, and $ b = \sqrt{e^2 – a^2}$ is the length of an

*.*

**imaginary axis of a hyperbola**Number $\varepsilon = \frac{e}{a}$ is called the * numeric eccentricity of a hyperbola*.

The last step to do is to draw the asymptotes. There will be two asymptotes with equations: $ y_1 = – \frac{b}{a}x$ and $ y_2 = \frac{b}{a}x$.

**How to constructing a hyperbola?**

Let’s say that we are given *foci* $ F_1, F_2$ and the length of real axis $2a$. We want to construct few points of a hyperbola from these information.

First, we’ll construct the center $O$, as a middle point of a line ${F_1 F_2}$.

Somewhere on the side we’ll draw a line $\overline{UV}$ whose length is equal to $2a$ and then extend it over $V$.

On that extension we’ll choose a point $ P_1$ and construct circles $ c_1 (F_1, \mid UP_1 \mid), c_2(F_2, \mid UP_1 \mid)$. The intersections of these circles will give us four different points of a hyperbola.

For more points you’ll use more other points on that extension, draw the circles and mark their intersections as new points.

**Hyperbola equation**

Just like in an ellipse, here we’ll also examine a hyperbola in a special position relating the coordinate system. We’ll set it in a way that the center of a hyperbola falls into the origin.

*Suppose $H$ is a hyperbola whose foci lie on the $x$- axis and its centre is the origin of a coordinate system. Then, the hyperbola is given with an equation:*

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

Where:

$ b^2 = e^2 – a^2$.

From this formula you’ll know the foci coordinates, $ F_1 (- e, 0)$, $ F_2 (e, 0)$ and the vertexes $ A(- a, 0)$, $ B ( a, 0)$.

Now you know almost everything you need to know in order to construct a hyperbola from its equation properly.

We already mentioned another important thing –* asymptotes*. Since $ y_1 = – \frac{b}{a}x$ and $ y_2 = \frac{b}{a}x$ you can directly find their equations.

**Example 1**. Draw a hyperbola with equation $\frac{x^2}{16} – \frac{y^2}{9} = 1$.

$ a = \sqrt{16} = 4$

$ b = \sqrt{9} = 3$

$ e^2 = a^2 + b^2 = 16 + 9 = 25 \rightarrow e = 5$

$ F_1= (-5, 0)$, $ F_2= (5, 0)$, $ A= (-4, 0)$, $ B= (4, 0)$

First asymptote: $ y_1 = -\frac{3}{4}x$.

Second asymptote: $ y_2 = \frac{3}{4}x$.

**What if the $x$ and $y$ variables changed their places?**

Let’s examine a hyperbola given with an equation $\frac{y^2}{16} – \frac{x^2}{9} = 1$

Now, the foci will be on the y – axis, and not on the x-axis. In other words, x – axis and y – axis changed their roles.

*Suppose $H$ is a hyperbola whose foci lie on $y$- axis and its center is the origin of the coordinate system.*

*Equation of that hyperbola is:*

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

*Or:*

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

This means that all the things we know how to calculate also change:

$ F_1 = (0, – e)$, $ F_2 = (e, 0)$, $ A (0, – a)$, $ B (0, a)$

This means that for a hyperbola $\frac{y^2}{16} – \frac{x^2}{9} = 1$ :

$ F_1 = (0, – 5)$, $ F_2 = (5, 0)$, $ A (0, – 4)$, $ B (0, 4)$

*A hyperbola whose real and imaginary axis are the same is called the isosceles hyperbola.*

**Isosceles hyperbola**

This means that $ a = b$ is valid equation.

$ E… \frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$

Or:

$ x^2 – y^2 = a^2$.

**Example 2**. Draw a hyperbola $ x^2 – y^2 = 16$

$ a = 4$, $ y_1 = x$, $ y_2 = – x$

**The condition to hyperbola and a line to meet**

If we want to algebraically determine an intersection of a line $ l… k = kx + l$, which is not an asymptote of a hyperbola, and a hyperbola $ H… $ $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$

$ y = kx + l$

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

If we insert $ kx + l$ instead of y into the other equation we’ll get:

$\frac{x^2}{a^2} – \frac{(kx + l)^2}{b^2} = 1$