Problem 44 - College Algebra

Question

Convert each equation to standard form by completing the square on \(x\) and \(y .\) Then graph the hyperbola. Locate the foci and find the equations of the asymptotes. $$4 x^{2}-y^{2}+32 x+6 y+39=0$$

Step-by-Step Solution

Answer

The hyperbola is graphed with center at (-4,3). The foci are at (-4 \pm 4\sqrt{5}, 3). The equations of the asymptotes are \(y= 2x+11\) and \(y= -2x+11\).

Step 1: Group \(x\) And \(y\)-terms

Rewrite the given equation \(4x^{2}-y^{2}+32x+6y+39=0\) as follows to group \(x\) and \(y\) terms: \(4(x^{2}+8x)-1(y^{2}-6y)=-39.\)

Step 2: Complete the Square

To complete the square, half the coefficient of the \(x\) and \(y\) terms, square it, and add it inside the parentheses. We have \(4[(x+4)^{2}-16] -1[(y-3)^{2}-9] = -39\). Now simplify to \(4(x+4)^{2} - 1(y-3)^{2} = 64\).

Step 3: Convert to Standard Form

To convert to standard form, divide through by 64 to get \((x+4)^{2} / 16 - (y-3)^{2} / 64 = 1\). Now this is in standard form of a hyperbola's equation.

Step 4: Locate the Center

From the standard form, the center of the hyperbola is \((-4,3)\). This can be obtained by taking the negative of the constant values present inside the brackets.

Step 5: Graph the Hyperbola

Plot the center, then plot points along the \(x\) and \(y\) axes 4 and 8 units away from the center. Draw the hyperbola using these plotted points as references.

Step 6: Find the Foci

To find the foci of the hyperbola, use the formula \(c=\sqrt{a^{2}+b^{2}}\), where \(a\) is the \(a\) value from the formula, and \(b\) is the \(y\)-denominator. Here, we get \(c=\sqrt{16+64}= \sqrt{80}=4 \sqrt{5}\). The foci are located at \((-4 \pm 4\sqrt{5}, 3)\) along the \(x\)-axis because the hyperbola opens left and right.

Step 7: Find the Equations of the Asymptotes

The equations of the asymptotes are given by \(\pm b/a(x-h)+k\), where \(h\) and \(k\) are the coordinates for the center of the hyperbola. This gives us the equations \(y-3= \pm 2(x+4)\) or \(y= \pm 2x+8+3\) or \(y= \pm 2x+11\).

Key Concepts

Completing the Square

Completing the square is a method used in algebra to form a perfect square trinomial from a quadratic equation. This process is beneficial when transforming equations into standard forms for different conic sections, such as a hyperbola.
To complete the square for a term like x² + bx:

• Take half of the coefficient b (the number in front of x), which is b/2.
• Square this result, giving you (b/2)².
• Add and subtract (b/2)² within the expression to maintain balance in the equation.

In the context of the hyperbola equation given, the terms' coefficients are adjusted to make the expressions inside the parentheses perfect squares. Completing the square is instrumental in leading an equation closer to its standard form, paving the way for further analysis and graphing.

Standard Form of a Conic Section

The standard form of a hyperbola is important because it reveals much about the graph's properties.It takes the general form:a hyperbola oriented left and right: \[ \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \]oriented up and down:\[ \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \]where (h, k) is the center of the hyperbola, a and b are the distances from the center to the vertices along the x and y axes, respectively.
In the problem at hand, converting from a general form to the standard form required completing the square and simplifying.The end result was:\[ \frac{(x+4)^2}{16} - \frac{(y-3)^2}{64} = 1 \]This form tells us that the hyperbola is centered at (-4, 3) and opens horizontally.

Asymptotes of a Hyperbola

Asymptotes are critical lines that the hyperbola approaches but never touches.They guide how the hyperbola's arms extend into infinity.For hyperbolas in standard form \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \), the asymptotes are figured by:y - k = ±(b/a)(x - h).
These lines give the hyperbola its characteristic 'X' shape.For our specific hyperbola, these asymptotes simplify to:\[ y = 2x + 11 \, \text{and} \, y = -2x + 11 \]Understanding the slopes of these lines can help in accurately sketching the hyperbola's arms, as they become increasingly closer to the asymptotes as you move further from the center.

Foci of a Hyperbola

The foci of a hyperbola are two fixed points, which this curve revolves around.They are an essential characteristic that offer insight into the eccentricity of the hyperbola.In math, the formula to find the distance to the foci from the center (\(c\)) uses:

• \(c = \sqrt{a^2 + b^2} \)

In our case, with \(a^2 = 16\) and \(b^2 = 64\), the foci are calculated as:\[ c = \sqrt{16 + 64} = 4\sqrt{5} \]This determines that the foci of our hyperbola are (-4 ± 4√5, 3), which are crucial for drawing the precise shape of the hyperbola.Focusing on these points aids in comprehending how the hyperbola stretches and proclaims its eccentricity.