Problem 18 - 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. \(4 x^{2}-8 x-9 y^{2}-72 y+112=0\)

Step-by-Step Solution

Answer

Standard form: \(\frac{(x-1)^2}{63} - \frac{(y+4)^2}{28} = 1\). Vertices: (1±\(\sqrt{63}\), -4); Foci: (1±\(\sqrt{91}\), -4); Asymptotes: \(y + 4 = \pm \frac{2}{3\sqrt{7}}(x - 1)\).

Step 1: Organize the Equation

Start by rearranging and grouping the terms: \[4x^2 - 8x - 9y^2 - 72y = -112\] Group the x terms together and the y terms together.

Step 2: Complete the Square

Complete the square for both grouped terms:1. For the x terms: - Factor out 4: \[4(x^2 - 2x)\] - Complete the square: \[4((x-1)^2 - 1)\] - Becomes: \[4(x-1)^2 - 4\]2. For the y terms: - Factor out -9: \[-9(y^2 + 8y)\] - Complete the square: \[-9((y+4)^2 - 16)\] - Becomes: \[-9(y+4)^2 + 144\]

Step 3: Simplify and Set into Standard Form

Substitute the completed squares back into the equation and simplify:\[4(x-1)^2 - 4 - 9(y+4)^2 + 144 = -112\]Combine constants, \[4(x-1)^2 - 9(y+4)^2 = -252\]Divide the entire equation by -252 to write it in standard form:\[\frac{(x-1)^2}{63} - \frac{(y+4)^2}{28} = 1\]

Step 4: Identify the Vertices and Foci

For the hyperbola \[\frac{(x-1)^2}{63} - \frac{(y+4)^2}{28} = 1\], the center is at (1, -4):- Semi-transverse axis a is \(\sqrt{63}\), vertices are at (1±\(\sqrt{63}\), -4)- Semi-conjugate axis b is \(\sqrt{28}\)- Distance to foci c: \(c = \sqrt{a^2 + b^2} = \sqrt{63 + 28} = \sqrt{91}\)- Foci are at (1±\(\sqrt{91}\), -4)

Step 5: Write Equations of Asymptotes

The equations for the asymptotes of a hyperbola \[\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\] are:\[y - k = \pm \frac{b}{a}(x - h)\]Plugging in h = 1, k = -4, a = \(\sqrt{63}\), b = \(\sqrt{28}\), we obtain:\[y + 4 = \pm \sqrt{\frac{28}{63}}(x - 1)\] or simplified:\[y + 4 = \pm \frac{2}{3\sqrt{7}}(x - 1)\]

Key Concepts

Vertices and Foci of a Hyperbola

Understanding the vertices and foci of a hyperbola is essential in graphing and interpreting the curve. The standard form of a hyperbola is \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] for a horizontal hyperbola. Here,

• \((h, k)\) is the center of the hyperbola.
• \(a\) is the distance from the center to each vertex on the x-axis.
• \(b\) is the distance from the center to each vertex along the y-axis.
• \(c\) is the distance from the center to each focus, calculated using the formula \(c = \sqrt{a^2 + b^2}\).

In this problem, the center is at (1, -4). The semi-transverse axis, \(a = \sqrt{63}\), gives the vertices at (1 ± \(\sqrt{63}\), -4). The foci, or focal points, are found at (1 ± \(\sqrt{91}\), -4), determined using \(c = \sqrt{91}\). Vertices and foci are key in showing the direction and 'spread' of the hyperbola.

Completing the Square for Hyperbola

Completing the square is crucial when transforming a quadratic equation into its standard form. It helps isolate squared terms to make the equation more manageable and understandable.To complete the square, follow these steps:

• Group x and y terms separately. In our exercise: \(4x^2 - 8x - 9y^2 - 72y = -112\).
• Factor out constants from each group, if needed. For x, factor out 4: \[ 4(x^2 - 2x) \]For y, factor out -9: \[ -9(y^2 + 8y) \].
• For each group, find what squares perfectly, adding and subtracting inside the parentheses to maintain equality. For x: \[ x^2 - 2x \] becomes \[ (x-1)^2 - 1 \].For y: \[ y^2 + 8y \] becomes \[ (y + 4)^2 - 16 \].
• Multiply back by the factors you pulled out, and adjust the equation accordingly.

This technique is instrumental in organizing complex quadratic terms into a clear and standard form of a hyperbola, like the final equation: \[ \frac{(x-1)^2}{63} - \frac{(y+4)^2}{28} = 1 \].

Asymptotes of Hyperbola

Asymptotes are lines the hyperbola approaches but never touches. They define the direction as the hyperbola extends towards infinity. Calculating the asymptotes helps in sketching an accurate graph of the hyperbola.The formula for the asymptotes of the standard form \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \] is given by:\[ y - k = \pm \frac{b}{a}(x - h) \].Substituting the values for our exercise:

• Center \((h, k) = (1, -4)\),
• "a" is equal to \(\sqrt{63}\),
• "b" is \(\sqrt{28}\),

The asymptote equations become:\[ y + 4 = \pm \sqrt{\frac{28}{63}} (x - 1) \] This formula simplifies into:\[ y + 4 = \pm \frac{2}{3\sqrt{7}} (x - 1) \].These lines give the boundaries of the hyperbola's branches and guide you in plotting its path. Understanding asymptotes provides insight into the hyperbola's shape and orientation.