Problem 17 - Algebra 2

Question

Find the foci of each hyperbola. Then draw the graph. $$ 36 x^{2}-8 y^{2}=288 $$

Step-by-Step Solution

Answer

The foci of the given hyperbola are at points \((2\sqrt{11}, 0)\) and \((-2\sqrt{11}, 0)\).

Step 1: Get given equation into standard hyperbola form

Divide both sides of the equation by 288, this gives: \((x^2/8) - (y^2/36) = 1\). Thus, \(a^2 = 8\) and \(b^2 = 36\). So, \(a = \sqrt{8}\), \(a = 2\sqrt{2}\) and \(b = 6\).

Step 2: Calculate the foci

Use the equation \(c = \sqrt{a^2 + b^2}\) to calculate c, in this case, \(c = \sqrt{(2\sqrt{2})^2 + (6)^2} = \sqrt{8 + 36} = \sqrt{44} = 2\sqrt{11}\). This says that the foci is at distances of \(2\sqrt{11}\) units to the right and left from the center of the hyperbola which is at the origin (0,0). Therefore, the foci points are \((2\sqrt{11}, 0)\) and \((-2\sqrt{11}, 0)\).

Step 3: Graph the hyperbola

Draw an xy-plane and plot the foci points. Then, sketch the hyperbola around these foci, keeping in mind that the hyperbola extends infinitely with asymptotes. Here, the given equation has a '+' on the \(x^2\) term and '-' on the \(y^2\) term, indicating it's a horizontal hyperbola squeezing along the x-axis. The center of hyperbola is at the origin, the and foci are at \((2\sqrt{11}, 0)\) and \((-2\sqrt{11}, 0)\). Draw the two branches of the hyperbola passing through the foci, opening rightward and leftward. Without a graph, this is textual description of how to do it.

Key Concepts

Foci of Hyperbola

The foci of a hyperbola are two special points located along its transverse axis. These points are crucial because the hyperbola is defined as the set of all points where the absolute difference of the distances to these foci is constant. For our given exercise, the equation is \(36x^2-8y^2=288\). First, we transform the equation into its standard form to identify the parameters that help us find the foci.
After obtaining the standard form \((x^2/8) - (y^2/36) = 1\), we identify \(a^2\) and \(b^2\). From here, we use the equation \(c = \sqrt{a^2 + b^2}\) to calculate the value of \(c\). This results in \(c = 2\sqrt{11}\), which tells us how far the foci are from the center.
These foci are located at \( (\pm 2\sqrt{11}, 0)\). The "\(\pm\)" symbol means there are two foci: one at \((2\sqrt{11}, 0)\) and the other at \((-2\sqrt{11}, 0)\). Having the foci plotted is crucial before attempting to sketch their corresponding hyperbola.

Standard Form of Hyperbola

The standard form of a hyperbola helps determine its basic characteristics, such as its shape, orientation, and key features like the center, vertices, and foci. For hyperbolas, the standard forms generally are:

• \((x^2/a^2) - (y^2/b^2) = 1\) for horizontally oriented hyperbolas
• \((y^2/a^2) - (x^2/b^2) = 1\) for vertically oriented hyperbolas

In our exercise, starting with the equation \(36x^2-8y^2=288\), converting it into standard form is crucial. This is achieved by dividing all terms by 288, leading us to \((x^2/8) - (y^2/36) = 1\). From this format, we learn that \(a^2 = 8\) and \(b^2 = 36\).
Knowing the standard form allows easy classification and simplifies the process of calculating other important features like the foci and asymptotes. Here, the larger denominator under \(y^2\) does not impact the orientation because the sign of terms in the equation implies it is horizontal, telling us how it will stretch between the axes.

Graphing Hyperbolas

Graphing hyperbolas can be straightforward with the correct approach and understanding of its features. Once we have the standard form, the next step is placing these features correctly on the graph. With the form \((x^2/8)-(y^2/36)=1\), it is apparent that this hyperbola stretches along the x-axis, being horizontal.
To begin graphing:

• Identify the center of the hyperbola, which is at the origin \( (0,0) \).
• Plot the calculated foci at \( (2\sqrt{11}, 0) \) and \((-2\sqrt{11}, 0) \).
• Determine the vertices, which are at \( (\pm \sqrt{8}, 0) \) or \( (\pm 2\sqrt{2}, 0) \).

Next, draw the asymptotes, which give a visual guide as to how the hyperbola opens; since the center is at the origin, they cross at (0,0) reflecting the hyperbola's bilateral symmetry. Finally, sketch the hyperbola passing through the vertices, ensuring the branches approach but never touch the asymptotes.
This visual representation helps one understand the behavior of the hyperbola and how it relates to the calculated foci.