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)$, centre (point $O$) and vertexes ($A$, $B$).
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:
$\frac{x^2}{a^2} – \frac{(kx + l)^2}{b^2} = 1$
