Problem 41 - College Algebra with Corequisite Support

Question

For the following exercises, sketch a graph of the hyperbola, labeling vertices and foci. \(64 x^{2}+128 x-9 y^{2}-72 y-656=0\)

Step-by-Step Solution

Answer

The hyperbola is centered at (-1,-4) with vertices at (-4,-4) and (2,-4), and foci near (-9.54,-4) and (7.54,-4).

Step 1: Rearrange the Equation

First, rearrange the given equation to collect the terms with the same variables together. Given the equation is \(64x^2 + 128x - 9y^2 - 72y - 656 = 0\). You organize the terms as \(64x^2 + 128x - 9y^2 - 72y = 656\).

Step 2: Complete the Square for x

To simplify the expression involving \(x\), factor out \(64\) from the \(x\) terms and complete the square. \[64(x^2 + 2x) = 64((x+1)^2 - 1)\].

Step 3: Complete the Square for y

Similarly, for \(y\) variables, factor \(-9\) and complete the square. \[-9(y^2 + 8y) = -9((y+4)^2 - 16)\].

Step 4: Rewrite the Equation

Substitute the completed squares back into the equation and simplify it to standard form: \[64((x+1)^2 - 1) - 9((y+4)^2 - 16) = 656\]Simplifying, \[64(x+1)^2 - 64 - 9(y+4)^2 + 144 = 656\]Combine constants, \[64(x+1)^2 - 9(y+4)^2 = 576\].

Step 5: Normalize the Equation

Divide every term by 576 to get the equation in standard form. \[\frac{(x+1)^2}{9} - \frac{(y+4)^2}{64} = 1\]. This is the equation of a hyperbola centered at \((-1,-4)\).

Step 6: Identify the Vertices and Foci

For the hyperbola \(\frac{(x+1)^2}{9} - \frac{(y+4)^2}{64} = 1\), vertices are along the transverse axis, given by \(a^2 = 9\), hence \(a=3\). Thus, the vertices are \((-1\pm3, -4)\) which are \((-1-3, -4)\) and \((-1+3, -4)\), i.e., \((-4, -4)\) and \( (2, -4)\).To find the foci, calculate \(c\) using \(c^2 = a^2 + b^2\) where \(b^2 = 64\), hence \(c = \sqrt{73}\).The foci are \((-1 \pm \sqrt{73}, -4)\).

Step 7: Sketch the Hyperbola

Based on the centers, vertices, and foci, draw the hyperbola on the coordinate plane:- Plot the center at \((-1,-4)\),- Mark the vertices at \((-4,-4)\) and \((2,-4)\),- Mark the foci approximately at \((-1-\sqrt{73}, -4)\) and \((-1+\sqrt{73}, -4)\), around \((-9.54,-4)\) and \((7.54,-4)\).- Symmetrically sketch the branches of the hyperbola opening left and right.

Key Concepts

Vertices

Vertices are key points of a hyperbola. They lie on the transverse axis, which is the line passing through the center of the hyperbola. For a horizontal hyperbola like \[\frac{(x+1)^2}{9} - \frac{(y+4)^2}{64} = 1\]the vertices are found at a distance of "a" units from the center. Here, \(a^2 = 9\) makes \(a = 3\).
Therefore, with the center at \((-1, -4)\), the vertices are \((-1-3, -4)\) and \((-1+3, -4)\).
This simplifies to \((-4, -4)\) and \((2, -4)\), confirming that they are positioned along the transverse axis, indicating where the hyperbola curves the widest.

Foci

The foci of a hyperbola are points within its branches, located on the same transverse axis as the vertices. The distance to these points helps define the shape of the hyperbola. To find the foci, you calculate \(c\) using the formula \(c^2 = a^2 + b^2\). Here, \(a^2 = 9\) and \(b^2 = 64\), so \(c = \sqrt{73}\), which is approximately 8.54.
Knowing the center is at \((-1, -4)\), you position the foci at \((-1-\sqrt{73}, -4)\) and \((-1+\sqrt{73}, -4)\).
This implies the foci are approximately \((-9.54, -4)\) and \((7.54, -4)\). These points are pivotal in making the hyperbola distinct, ensuring its focus-driven shape.

Conic Sections

Conic sections are curves obtained by slicing a cone with a plane. Hyperbolas are one of these fascinating shapes, alongside ellipses, parabolas, and circles.

• They arise when the intersection occurs at an angle parallel to the cone's axis, forming two distinct, open curves.
• The unique geometry of a hyperbola stretches out from the center, with each curve reflecting the properties defined by its equation.

For instance, the hyperbola's equation given in the problem originates from rearranging and simplifying the conic section's general second-degree equation. Understanding these shapes provides insight into real-world applications, such as satellite navigation systems, where hyperbolas aid in determining precise locations.

Graphing Hyperbolas

Graphing a hyperbola involves visualizing its shape, which distinctively opens outward in both directions from a central vertex. Here’s how to graphically represent it:

• First, identify the center of the hyperbola. For our equation, the center is located at \((-1, -4)\).
• Mark the vertices. In our case, these are \((-4, -4)\) and \((2, -4)\).
• Plot the foci, which are critical in sketching the correct shape, at approximately \((-9.54, -4)\) and \((7.54, -4)\).
• Using these plotted points, draw two curves that mirror each other around the center and pass close to the foci.

Adjust the graph until the hyperbola correctly aligns with its defining points. Understanding this aids not only in solving equations but also in comprehending the significance of its open, intersecting design, relevant in fields like physics and engineering.