Problem 74 - A Graphical Approach to Precalculus with Limits

Question

Find an equation for each hyperbola. Vertices \((5,-2)\) and \((1,-2) ;\) asymptotes \(y=\pm \frac{3}{2}(x-3)-2\)

Step-by-Step Solution

Answer

Equation: \(\frac{(x-3)^2}{4} - \frac{(y+2)^2}{9} = 1\). Center: \((3, -2)\). Orientation: horizontal.

Step 1: Determine center of the hyperbola

The center of the hyperbola lies at the midpoint of the vertices. The vertices given are \((5,-2)\) and \((1,-2)\). Thus, the center \((h,k)\) can be calculated as follows:\[\left( \frac{5+1}{2}, \frac{-2-2}{2} \right) = (3, -2)\].So, the center of the hyperbola is \((3, -2)\).

Step 2: Identify the orientation and equation form

Since the vertices \((5,-2)\) and \((1,-2)\) have the same y-coordinate, the transverse axis is horizontal. Therefore, the equation of the hyperbola is of the form:\[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\]where \((h, k) = (3, -2)\).

Step 3: Calculate 'a' using the distance between vertices

The distance between the vertices of a hyperbola is \(2a\). The distance between the given vertices \((5, -2)\) and \((1, -2)\) is \[\sqrt{(5-1)^2 + (-2-2)^2} = \sqrt{16} = 4\]Therefore, \(2a = 4\) which means \(a = 2\).

Step 4: Calculate 'b' using the slope of asymptotes

The slope of the asymptotes for a hyperbola is \(\pm \frac{b}{a}\). The given slope of the asymptotes is \(\pm \frac{3}{2}\). Therefore, \[\frac{b}{2} = \frac{3}{2} \implies b = 3.\]

Step 5: Write the equation of the hyperbola

Substitute the center \((h, k)\), \(a\), and \(b\) into the hyperbola equation form:\[\frac{(x-3)^2}{2^2} - \frac{(y+2)^2}{3^2} = 1\]Thus, the equation of the hyperbola is: \[\frac{(x-3)^2}{4} - \frac{(y+2)^2}{9} = 1.\]

Key Concepts

Equation of Hyperbola

A hyperbola is a type of conic section that looks like two mirrored, open curves.
The equation of a hyperbola in its standard form can be determined by its orientation, which depends on the position of its vertices.For a hyperbola with a horizontal transverse axis, the equation follows this format:
\[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]If the transverse axis is vertical, the equation switches to:
\[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]In our exercise, because the vertices differ only in the x-coordinate, the transverse axis is horizontal.
So, the standard equation becomes:

• \( (h, k) \) is the center of the hyperbola.
• \( a \) is the half-distance between vertices.
• \( b \) is calculated from the asymptotes' slope.

These parameters insert into the hyperbola's equation to characterize its precise shape in the coordinate plane.

Vertices

Vertices of a hyperbola are the points that lie on the transverse axis in the widest part of each branch of the hyperbola.
For any hyperbola, these points are essential in defining its size and shape.The vertices provided in our problem are \((5, -2)\) and \((1, -2)\).
They both share the same y-coordinate, meaning the transverse axis is horizontal in this particular hyperbola.
This determines which equation form of the hyperbola to use, and it also defines the distance \(2a\):

• Measure the distance between vertices to find \(2a\).
• Distance calculated using distance formula: \( \sqrt{(5-1)^2 + (-2-(-2))^2} = \sqrt{16} = 4 \).
• So, \(a = 2\).

Knowing \(a\) helps plot the vertices and guides the complete shape of the hyperbola.

Asymptotes

Asymptotes of a hyperbola are the lines that the branches of the hyperbola approach but never actually meet.
For hyperbolas, they appear as diagonal lines intersecting through the center, hinting at the general direction of the curves.In the given exercise, the asymptotes have been given as:
\( y = \pm \frac{3}{2}(x - 3) - 2 \).The general form for asymptotes of a horizontally oriented hyperbola is:
\[ y - k = \pm \frac{b}{a}(x - h) \]

• \(\pm \frac{3}{2}\) is the slope indicating \(\frac{b}{a}\).
• Setting \(a = 2\), we solve \(\frac{b}{2} = \frac{3}{2}\) which shows \(b = 3\).

Asymptotes guide us by showing the line along which the hyperbola branches open, further defining its extended behavior.

Hyperbola Center

The center of a hyperbola is the midpoint between its vertices and marks the point of symmetry for its shapes.Found by taking the average of the x- and y-coordinates of the vertices, the center defines the equation's form:
The midpoint formula gives us:

• \( (h, k) = \left( \frac{5+1}{2}, \frac{-2-2}{2} \right) = (3, -2) \)

The center \((3, -2)\) acts as a guiding point, tying together the hyperbola's vertex alignment and its asymptotes.
Starting from this central location, the hyperbola expands outward in its curves, maintaining symmetrical beauty across its center.