Given two points \(F_1\) and \(F_2\) in the plane lying a distance \(2c\) apart and given a distance \(2a < 2c\), the set of points whose distances to \(F_1\) and \(F_2\) respectively differ by \(2a\) is a hyperbola.

F1 and F2 are the *foci* of the hyperbola.

You will need a compass and a blank sheet of paper.

On a horizontal dotted line, mark two points \(F_1\) and \(F_2\) which will be the foci of the hyperbola.

Construct a dotted perpendicular line to \(F_1F_2\) through the midpoint of that segment. Label the midpoint as the center.

Place the point of the compass at the center and open the compass to a fixed radius less than the distance from the center to the foci.

Mark points \(V_1\) and \(V_2\) where the pencil of the compass crosses the line that passes through \(F_1\) and \(F_2\). These are the vertices of the hyperbola.

Construct vertical dotted lines through \(V_1\) and \(V_2\) .

Place the point of the compass at the center and open it to a radius equal to the distance from the center to either focus. This will be denoted as the distance \(c\).

Mark the four points on the verticals through\(V_1\) and \(V_2\) where the pencil of the compass crosses.

Connect those four points into a dotted rectangle. This is the *central rectangle* of the hyperbola.

Using dotted lines, extend the diagonals of that rectangle. These are the two *asymptote*s of the hyperbola.

From each of the vertices \(V_1\) and \(V_2\), draw a line curving away from the center and toward each of the asymptotes.

The hyperbola is a graph which is in two separate pieces. The line through the foci is the transverse axis (`transverse' is Latin for "goes across) and the line through the center perpendicular to the transverse axis is the *conjugate* axis.

Using the hyperbola that you drew in Exercise 9.3.1, denote the distance from the center \(C\) to \(V_1\) and \(C\) to \(V_2\) as \(a\).

Find the difference of the distances from \(V_1\) to \(F_2\) and \(V_1\)to \(F_1\) in terms of \(a\).

Do the same for the sum of the distances from \(V_2\) to \(F_1\) and \(V_2\) to \(F_2\).

The width of the central rectangle you drew in Exercise 9.3.1 is \(2a\), and the length of its diagonals is \(2c\). Given that its height is \(2b\), find the relationship between \(a\), \(b\) and \(c\).

The major axis contains the two foci of a hyperbola and can be any line in the plane, but we will consider only those hyperbolas whose transverse axis is either horizontal or vertical.

Let us begin by considering a horizontal hyperbola with center at the origin and foci \(( \pm c, 0 )\) on the \(x\)-axis. Then the \(x\)-axis is the major axis. Let \(( \pm a, 0 )\) denote the vertices and \(( 0, \pm b )\) the points where the central rectangle crosses the \(y\)-axis. The line constructed through the points \((0,\pm b)\) is called the *conjugate axis*. Construct the central rectangle as you did in Exercise 9.3.1. The distance from the center \(( 0, 0 )\) to the corners of the central rectangle is the same as the distance to the foci \(( \pm c, 0 )\). The equation of this `horizontal' hyperbola is

\[\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\]

Interesting fact: If we draw an *ellipse* inside the central rectangle with vertices \((\pm a,0)\) and \((0,\pm b)\) then it has the complementary equation \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\).

If the basic hyperbola is shifted \(h\) units horizontally and \(k\) units vertically so that its center is at \((h,k)\) rather than at the origin, the resulting hyperbola will have equation

\[\frac{(x-h)^2}{a^2}-\frac{(y-k)^2}{b^2}=1\]

Next, let us consider a vertical hyperbola with center at the origin and foci \(( 0, \pm c )\) on the y-axis. Then the \(y\)-axis is the transverse axis. Let \(( 0, \pm a )\) denote the vertices and \(( \pm b, 0 )\) the points where the central rectangle crosses the \(x\)-axis. Construct the central rectangle. The distance from the center ( 0, 0 ) to the corners of the central rectangle is the same as the distance to the foci. The equation of this â€˜verticalâ€™ hyperbola is

\[\frac{y^2}{a^2}-\frac{x^2}{b^2}=1\]

If this hyperbola is shifted \(h\) units horizontally and \(k\) units vertically, the resulting ellipse will have equation

\[\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}=1\]

In either case \(c^2 = a^2 + b^2\).

The following diagram shows the graphs of the following five equations:- \(\dfrac{(x-1)^2}{9}-\dfrac{(y+2)^2}{4}=1\) in black
- \(\dfrac{(y+2)^2}{9}-\dfrac{(x-1)^2}{4}=1\) in red
- \(\dfrac{(x-1)^2}{9}+\dfrac{(y-2)^2}{4}=1\) in blue
- \((x-1)^2+(y+2)^2=3^2+2^2\) Circle
- \(\dfrac{x-1}{3}-\dfrac{y+2}{2}=1\) Increasing asymptote
- \(\dfrac{x-1}{3}+\dfrac{y+2}{2}=1\) Decreasing asymptote

Find the equation of the hyperbola with center \(( 0, 0 )\), vertex \(( -3, 0 )\) and focus \(( 5, 0 )\) and draw the graph.

Find the equation of the hyperbola with center \(( 3, -2 )\), vertex \(( 3, -3 )\) and focus \(( 3, 0 )\) and draw the graph.

Complete the square to put the equation into standard form, then find the center, vertices and foci.

\(4x^2 - y^2 + 16x + 2y + 19 = 0\)