Problem 40 - Algebra and Trigonometry Real Mathematics, Real People

Question

Find the standard form of the equation of the hyperbola, (b) find the center, vertices, foci, and asymptotes of the hyperbola, and (c) sketch the hyperbola. Use a graphing utility to verify your graph. 9 x^{2}-y^{2}+54 x+10 y+55=0

Step-by-Step Solution

Answer

The standard form of the equation of the hyperbola is \(\frac{(x+3)^2}{9} - \frac{(y-5)^2}{16} = 1\). The center of the hyperbola is at \(-3, 5\), vertices are at \((-6,5)\) and \((0,5)\), foci are at \((-8,5)\) and \((2,5)\), and the equations of the asymptotes are \(y = 5 \pm \frac{4}{3}(x + 3)\).

Step 1: Write the equation in standard form

We first group the \(x\) terms and the \(y\) terms, then complete the square for both \(x\) and \(y\). The given equation is \(9x^{2}+54x-y^{2}+10y+55=0\). This translates into the standard form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\). Completing the square we get \(9(x^2+6x+9) - (y^2-10y+25) = 55 - 9*9 + 25\). Hence, our equation in standard form is \(\frac{(x+3)^2}{9} - \frac{(y-5)^2}{16} = 1\).

Step 2: Identify the center, vertices, foci and asymptotes

From the standard form, we can identify the center \((h, k)\) is \(-3, 5\). Here, \(a=3\) and \(b=4\). Since the \(x\)-term possesses the positive sign in the equation, the hyperbola is horizontal. Thus, the vertices are \(h \pm a, k\), which means the vertices are at \((-3 \pm 3, 5)\) or at points \((-6,5)\) and \((0,5)\). The foci will be at \(h \pm c, k\), where \(c\) is the distance from the center to the focus points. We calculate \(c\) as \(\sqrt(a^2+b^2) = \sqrt(9+16) = 5\). Hence, the foci are at points \((-8,5)\) and \(2,5)\). The equations of the asymptotes for a horizontal hyperbola are given by \(y = k \pm \frac{b}{a}(x - h)\). Therefore, the equations of asymptotes are \(y = 5 \pm \frac{4}{3}(x + 3)\).

Step 3: Sketch the hyperbola

First plot the center at \(-3, 5\). Plot the vertices and Foci. Draw a rectangle around the vertices and draw the diagonals of the rectangle, which represent the asymptotes. Bounded by the asymptotes, sketch the two branches of the hyperbola going through the vertices. With this, the sketch of the hyperbola is complete.

Key Concepts

Standard Form

The standard form of a hyperbola is crucial for identifying its key features such as the center, vertices, and foci. The equation of a hyperbola can be written in the standard form as:

• For a horizontal hyperbola: \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \)
• For a vertical hyperbola: \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \)

Here, \((h, k)\) represents the center of the hyperbola. The terms \(a\) and \(b\) are values that determine the distances related to the vertices and the asymptotes. In the given problem, by completing the square on \(x\) and \(y\), we find the standard form as: \( \frac{(x+3)^2}{9} - \frac{(y-5)^2}{16} = 1 \).
This indicates that the hyperbola is oriented horizontally, or along the x-axis.

Center

The center of a hyperbola is a key reference point from which you determine other elements like vertices and foci. From the standard form equation \( \frac{(x+3)^2}{9} - \frac{(y-5)^2}{16} = 1 \), the values of \(h\) and \(k\) determine the center. Here, the center is at the point \((-3, 5)\).
The center can be thought of as the midpoint between the hyperbola's vertices and also the midpoint between its foci. It's important to locate the center correctly because it acts as the balance point of the hyperbola.
Once you have your center, you can map out the hyperbola's structure more easily by measuring outward to find vertices and focus points.

Vertices

Vertices are the points on the hyperbola that are closest to its focus. For a hyperbola, the vertices lie along the transverse axis, which correlates with whether the hyperbola is horizontal or vertical. Since our given hyperbola has the horizontal form, we find its vertices using the formula:
\(h \pm a, k\).
From our standard equation, the vertices are \((-3 \pm 3, 5)\), which gives:

• \((-6, 5)\)
• \((0, 5)\)

Vertices mark the start and end of the traversing part of the hyperbola, the arcs that get infinitely closer to the asymptotes. They reflect the extent to which the hyperbola "opens" along the axis of symmetry.

Foci

Foci (plural of focus) are crucial for understanding the hyperbola's geometry. Like the center and vertices, knowing the foci positions helps map out the entire hyperbola. The foci are situated along the transverse axis, further out from the center than the vertices.
The distance from the center to a focus point is given by \(c = \sqrt{a^2 + b^2} \). In this case, calculating \(c\) yields a value of 5. Therefore, the foci are positioned at \((-3 \pm 5, 5)\), which gives us the points:

• \((-8, 5)\)
• \((2, 5)\)

The concept of the foci can be thought of in terms of the hyperbola's reflection properties, playing a critical role in the shape's definition.

Asymptotes

Asymptotes are the lines that the hyperbola approaches but never touches. For hyperbolas, asymptotes provide a boundary framework for sketching the hyperbola's shape and are particularly useful for predicting its behavior at infinity.
The equations of the asymptotes for a horizontal hyperbola centered at \((h, k)\) are given as:

• \(y = k \pm \frac{b}{a}(x - h)\)

Using this formula for our hyperbola, the asymptote equations become \(y = 5 \pm \frac{4}{3}(x + 3)\). These lines are where each arm of the hyperbola heads off to infinity, lying at an angle determined by the ratio \(\frac{b}{a}\), thus dictating the slope of each asymptote.