Problem 76 - Algebra 2

Question

Find the foci of each hyperbola. Draw the graph. $$ \frac{x^{2}}{36}-\frac{y^{2}}{49}=1 $$

Step-by-Step Solution

Answer

The foci of the hyperbola \(\frac{x^{2}}{36}-\frac{y^{2}}{49}=1\) are located at \((\pm \sqrt{85}, 0)\).

Step 1: Identify the values of a and b

From the given equation \(\frac{x^{2}}{36}-\frac{y^{2}}{49}=1\) we see that \(a^2=36\) and \(b^2=49\). Therefore, \(a = 6\) and \(b = 7\).

Step 2: Calculate the distance of foci from the center

The formula to calculate the distance of foci from the center in case hyperbola opens horizontally is \(c=\sqrt{a^2+b^2}\). Substituting the values, we get \(c=\sqrt{6^2+7^2}=\sqrt{36+49}=\sqrt{85}\). So the coordinates of the foci will be \((\pm c, 0)\), or \((\pm \sqrt{85}, 0)\).

Step 3: Draw the graph of the hyperbola

Sketch an xy-coordinate plane. The center of the hyperbola is at the origin (0,0). It opens left and right with vertices at \((\pm a, 0)\) or \((\pm 6, 0)\). Draw two points for the foci at \((\pm \sqrt{85}, 0)\), and then sketch in the hyperbola, making sure it opens towards and away from the foci.

Key Concepts

Foci of a Hyperbola

In a hyperbola, the foci are two fixed points used to define the curve. For our given equation, \(\frac{x^2}{36} - \frac{y^2}{49} = 1\), the hyperbola opens horizontally. This implies that the foci are aligned horizontally along the x-axis. The formula to find the distance from the center to the foci is \(c = \sqrt{a^2 + b^2}\). This is derived from the Pythagorean relationship specific to hyperbolas. Plugging in the given values, \(a^2 = 36\) and \(b^2 = 49\), we calculate \(c = \sqrt{36 + 49} = \sqrt{85}\). Thus, the foci are located at \((\pm \sqrt{85}, 0)\). These foci represent points towards which the hyperbola curves and are critical for understanding the hyperbola's shape and direction.

Graphing Hyperbolas

Graphing a hyperbola involves several steps. Begin by identifying the center of your hyperbola, which in this case is the origin at \((0,0)\). For the equation \(\frac{x^2}{36} - \frac{y^2}{49} = 1\), the vertices are positioned at \((\pm 6, 0)\) because \(a = 6\). Place these points on the x-axis, as our hyperbola opens horizontally.
Next, plot the foci calculated earlier at \((\pm \sqrt{85}, 0)\). These added points help illustrate how the hyperbola curves towards and away from them. Once you've marked the vertices and foci, sketch the hyperbola making sure it opens outwards from the vertices, creating two branches that move away towards the foci.

Hyperbola Equation

The equation of a hyperbola in standard form expresses how the hyperbola relates to its axes. For horizontal-opening hyperbolas, the equation takes the form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Here, \(a\) and \(b\) help determine the vertices and the co-vertices of the hyperbola. Specifically, \(a\) is the distance from the center to each vertex along the x-axis, and \(b\) relates to how tight or wide the branches appear when centered around \((0,0)\).
The equation \(\frac{x^2}{36} - \frac{y^2}{49} = 1\) reveals \(a = 6\) and \(b = 7\), dictating the hyperbola's width and height, respectively. Analyzing these values helps in sketching, locating foci, and understanding the overall behavior of the hyperbola.