Problem 19 - College Algebra

Question

Write the equation for the hyperbola in standard form if it is not already, and identify the vertices and foci, and write equations of asymptotes. \(-9 x^{2}-54 x+9 y^{2}-54 y+81=0\)

Step-by-Step Solution

Answer

The equation is \(\frac{(x+3)^2}{9} - \frac{(y+3)^2}{9} = 1\); vertices: (-6, -3), (0, -3); foci: (-3±4.24, -3); asymptotes: \(y=x-6\), \(y=-x\).

Step 1: Rearrange Terms

Rewrite the given equation grouping the x and y terms together. The terms are: \(-9x^2 - 54x + 9y^2 - 54y = -81\).

Step 2: Factor Coefficients

Factor out the common factor from each group. For x terms: \(-9(x^2 + 6x)\) and for y terms: \(+9(y^2 + 6y)\).

Step 3: Complete the Square

Complete the square inside each parenthesis. For the x-group: \(x^2 + 6x\) becomes \((x+3)^2 - 9\). For the y-group: \(y^2 + 6y\) becomes \((y+3)^2 - 9\).

Step 4: Substitute Back and Simplify

Substitute the completed squares back into the equation and simplify: \(-9((x+3)^2 - 9) + 9((y+3)^2 - 9) = -81\). Which simplifies to: \(-9(x+3)^2 + 81 + 9(y+3)^2 - 81 = -81\).

Step 5: Simplify Further

Combine the constant terms and rearrange: \(-9(x+3)^2 + 9(y+3)^2 = -81\). Then divide through by \(-81\) to get the standard form of the hyperbola: \(\frac{(x+3)^2}{9} - \frac{(y+3)^2}{9} = 1\).

Step 6: Identify Vertices and Foci

Vertices are at (-3 ± 3, -3) or (-6, -3) and (0, -3). The foci are located at (-3 ± √18, -3) or (-3 ± 4.24, -3).

Step 7: Equations of Asymptotes

Since this is a horizontal hyperbola, the equations of asymptotes are of the form \(y = k ± b/a (x - h)\) which becomes \(y = -3 ± 1/1 (x + 3)\), leading to \(y = x - 6\) and \(y = -x\).

Key Concepts

Standard Form Equation

The standard form equation of a hyperbola helps us understand its orientation and essential properties. Generally, a hyperbola can be expressed in two fundamental formulas based on its orientation:

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

For the given equation, \(-9x^2 - 54x + 9y^2 - 54y + 81 = 0\), we rearrange and simplify it, ultimately arriving at the standard form \(\frac{(x+3)^2}{9} - \frac{(y+3)^2}{9} = 1\). Since the \(x\) term is positive, this confirms that our hyperbola is horizontal.

Vertices

Vertices are crucial points on a hyperbola that provide insights into its width and location. They are situated at \((h \pm a, k)\) for a horizontal hyperbola and \((h, k \pm a)\) for a vertical one. From the standard form \(\frac{(x+3)^2}{9} - \frac{(y+3)^2}{9} = 1\), we identify that \(a^2 = 9\), which means \(a = 3\).
Here, the center of the hyperbola is \((-3, -3)\). So the vertices of the hyperbola are found at \((-3 \pm 3, -3)\) or \((-6, -3)\) and \((0, -3)\). These points symbolize how far the hyperbola stretches along its primary axis, which in this case, is horizontal.

Foci

The foci of a hyperbola are pivotal points inside its branches that help define its shape. They are determined by the formula \((h \pm c, k)\) for horizontal hyperbolas and \((h, k \pm c)\) for vertical ones, where \(c\) is obtained from \(c^2 = a^2 + b^2\).
In our scenario, both \(a^2\) and \(b^2\) are 9, making \(c^2 = 9 + 9 = 18\). Therefore, \(c = \sqrt{18}\) or approximately \(4.24\). Given the center \((-3, -3)\), the foci are at the points \((-3 \pm \sqrt{18}, -3)\), which equates to approximately \((-3 \pm 4.24, -3)\). These reveal how closely the branches come to converging near the center.

Asymptotes

Asymptotes are crucial for understanding the general behavior and shape of the hyperbola's branches. They represent lines that the branches of the hyperbola approach but never actually meet. For a hyperbola, these lines pass through the center and their equations take the general form:

• Horizontal hyperbola: \(y = k \pm \frac{b}{a}(x - h)\)
• Vertical hyperbola: \(x = h \pm \frac{a}{b}(y - k)\)

In this example, since it is a horizontal hyperbola, we use \(y = k \pm \frac{b}{a}(x - h)\). Plugging in the values gives us \(y = -3 \pm \frac{1}{1}(x + 3)\). This results in the equations of the asymptotes: \(y = x - 6\) and \(y = -x\). These asymptotes guide the hyperbola's shape and trajectory as its branches extend indefinitely.

Complete the Square

Completing the square simplifies quadratic expressions, making it easier to convert them into a standard form. This technique involves creating a perfect square trinomial from a quadratic expression.
Starting with the given expression \(-9x^2 - 54x + 9y^2 - 54y = -81\), we rearrange and factor the terms, leading to the forms \(-9(x^2 + 6x)\) and \(+9(y^2 + 6y)\). To complete the square for each component:

• For \(x\), transform \(x^2 + 6x\) to \((x+3)^2 - 9\)
• For \(y\), transform \(y^2 + 6y\) to \((y+3)^2 - 9\)

Substituting these back into the equation results in a form easily rearranged into the standard hyperbola equation. This method not only simplifies the algebra but also reveals the hyperbola's center, vertices, and other vital characteristics.