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Hyperbolas Solutions

A hyperbola is a conic section that is an open curve with two symmetric u-shaped branches set between asymptotes.
A hyperbola is called a deep verticle slice when referring to conic sections.
The equation of a hyperbola has two general forms to represent either a horizontal or vertical graph.
The hyperbola is broken into two parts, or branches, that fit between two slant asymptotes that meet at the center, $(h, k)$ .

Note

Recall from earlier lessons that an asymptote is a line that a graph approaches as it moves toward infinity. An asymptote is an invisible guideline that sets the boundaries of the graph. This means that it is not always drawn.

The asymptotes of hyperbolas will be written in point-slope form where $m = \pm \frac{b}{a}$ .

	Horizontal	Vertical
General Form	$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$	$\frac{{(y - k)}^{2}}{b^{2}} - \frac{{(x - h)}^{2}}{a^{2}} = 1$
Asymptotes
Branches	left and right $(h, k)$	above and below $(h, k)$
Transverse Axes	horizontal	vertical

The branches of the hyperbola intersect the transverse axis at the vertices . 
The vertices and co-vertices form a rectangle that can be used to graph the slant asymptotes diagonally across the rectangle and through the center, $(h, k)$ . 
The direction that the branches open is determined by the form of the equation. 
Note the direction on your paper with a quick sketch so that you can confirm your final graph matches it.

Example 1

Mark the center, asymptotes, and the vertices and co-vertices on the graph for the equation:   $\frac{{(x - 1)}^{2}}{25} - \frac{{(y + 3)}^{2}}{4} = 1$

Plan
Determine the center, a, b 
Mark the vertices and co–vertices 
Mark the asymptotes 
Write the equation of the asymptotes

Center $(1, - 3)$
$a = 5 b = 2$

$m = \pm \frac{b}{a} m = \pm \frac{2}{5}$

$y + 3 = \frac{2}{5} (x - 1) y + 3 = - \frac{2}{5} (x - 1)$

Example 2

Mark the center, asymptotes, and the vertices and co-vertices on the graph for the equation: 
$\frac{y^{2}}{9} - \frac{x^{2}}{36} = 1$

Center $(0, 0)$

$a = 6 b = 3$

$m = \pm \frac{b}{a} m = \pm \frac{3}{6} = \pm \frac{1}{2}$

$y = \frac{1}{2} x y = - \frac{1}{2} x$

Note

When the center is $(0, 0)$ , it is not necessary to write the equation in point-slope form.