Problem 29 - College Algebra with Corequisite Support

Question

For the following exercises, find the equations of the asymptotes for each hyperbola. \(9 x^{2}-18 x-16 y^{2}+32 y-151=0\)

Step-by-Step Solution

Answer

The asymptotes are \(y = \frac{3}{4}x - \frac{1}{4}\) and \(y = -\frac{3}{4}x + \frac{7}{4}\).

Step 1: Rearrange the equation

Start by rearranging the given equation in the standard form of a hyperbola. The given equation is \(9x^2 - 18x - 16y^2 + 32y - 151 = 0\). We will group the x terms and y terms together: \(9(x^2 - 2x) - 16(y^2 - 2y) = 151\).

Step 2: Complete the square for x terms

To complete the square for the x terms, take the coefficient of x (which is -2), halve it to get -1, and square it to get 1. Add and subtract 1 inside the brackets: \(9((x-1)^2 - 1) - 16(y^2 - 2y) = 151\). Simplify it to \(9(x-1)^2 - 9 - 16(y^2 - 2y) = 151\).

Step 3: Complete the square for y terms

For the y terms, take the coefficient of y (which is -2), halve it to get -1, and square it to get 1. Add and subtract 1: \( -16((y-1)^2 - 1) = -16(y^2 - 2y + 1 - 1) \). Simplify it to \(-16(y-1)^2 + 16 = 0\).

Step 4: Simplify and solve

Combine all terms having rearranged the equation: \(9(x-1)^2 - 16(y-1)^2 = 151 + 9 - 16\). This simplifies to \(9(x-1)^2 - 16(y-1)^2 = 144\). Divide the entire equation by 144 to match the standard form: \(\frac{(x-1)^2}{16} - \frac{(y-1)^2}{9} = 1\).

Step 5: Identify the asymptotes

For hyperbolas of the form \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the asymptotes are given by \(y - k = \pm\frac{b}{a}(x - h)\). Here, \(h = 1, k = 1, a^2 = 16, b^2 = 9, a = 4, b = 3\). Therefore, the asymptotes are given by the equations \(y - 1 = \pm\frac{3}{4}(x - 1)\).

Step 6: Write final equations of asymptotes

Plug in \(h\) and \(k\) into the asymptote equation: \(y - 1 = \pm\frac{3}{4}(x - 1)\). Thus the equations of the asymptotes are \(y = \frac{3}{4}(x-1) + 1\) and \(y = -\frac{3}{4}(x-1) + 1\), which can simplify to \(y = \frac{3}{4}x - \frac{1}{4}\) and \(y = -\frac{3}{4}x + \frac{7}{4}\).

Key Concepts

Completing the Square

Completing the square is an invaluable technique in algebra. It simplifies expressions to reveal the structure of conic sections like hyperbolas. When given a quadratic, the goal is often to transform it into a perfect square trinomial, which is an expression of the form \( (x - h)^2 \).
To complete the square:

• Identify the quadratic and linear terms of the variable, such as \(x^2\) and \(x\).
• Take the coefficient of the linear term, divide it by 2, then square it.
• Add and subtract this squared number within the expression.
• Rearrange the resulting expression into the form of a square trinomial.

By completing the square for both \(x\) terms and \(y\) terms, we transform the given equation of a hyperbola into its standard form. This process makes finding the equations of asymptotes more straightforward.

Standard Form of a Hyperbola

The standard form of a hyperbola reveals much about its characteristics, particularly the asymptotes' direction and orientation. For a hyperbola centered at \( (h, k)\), it can have either of these forms:

• Horizontal transverse axis: \((x-h)^2/a^2 - (y-k)^2/b^2 = 1\)
• Vertical transverse axis: \((y-k)^2/b^2 - (x-h)^2/a^2 = 1\)

In our problem, the hyperbola takes the form \( \frac{(x-1)^2}{16} - \frac{(y-1)^2}{9} = 1\), indicating a horizontal transverse axis. This form reveals that:

• The center of the hyperbola is at \( (1, 1)\).
• The values \(a^2 = 16\) and \(b^2 = 9\) tell us about the distances related to the vertices and asymptotes.

Converting the general quadratic equation to this standard form involves rearranging and simplifying through completion of the square.

Equations of Asymptotes

Asymptotes are straight lines that a hyperbola approaches but never touches. They provide a snapshot of the hyperbola's "direction." For the specific form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the slopes of the asymptotes are determined by \(\pm \frac{b}{a}\).

• Calculate the slopes: \( \frac{b}{a} = \frac{3}{4}\)
• Use the center of the hyperbola to express asymptote equations: \( y - k = \pm \frac{3}{4}(x - h)\).

In our example, these equations convert to \( y - 1 = \pm \frac{3}{4}(x - 1)\), leading to the final expressions:

• \( y = \frac{3}{4}x - \frac{1}{4}\)
• \( y = -\frac{3}{4}x + \frac{7}{4} \)

These equations define lines intersecting at the hyperbola's center, framing its open branches.

Coordinate Geometry

Coordinate geometry, or analytic geometry, plays a central role in connecting algebra with geometric curves. When investigating shapes like hyperbolas, we link equations to the visual structure by plotting points and drawing curves on a coordinate plane.

• Centers, vertices, and asymptotes are plotted from calculated points.
• Equations involving \(x\) and \(y\) help bridge algebraic expressions with geometric representations.

Our hyperbola \( \frac{(x-1)^2}{16} - \frac{(y-1)^2}{9} = 1\) informs us that its center is at \( (1, 1)\).
The asymptotes provide direction, sloping at \( \pm \frac{3}{4}\). With a clear understanding of these aspects, coordinate geometry helps visualize how a hyperbola 'opens' and behaves relative to its axes and asymptotes.