Problem 15 - Algebra 2

Question

Find the foci of each hyperbola. Then draw the graph. $$ \frac{x^{2}}{36}-\frac{y^{2}}{169}=1 $$

Step-by-Step Solution

Answer

The foci of the hyperbola are at (-sqrt{205}, 0) and (sqrt{205}, 0).

Step 1: Identify the a and b values from the equation

From the equation \(\frac{x^{2}}{36}-\frac{y^{2}}{169}=1\), we can see that it is in the form \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\). This implies a hyperbola oriented along the x-axis. Its semi-major axis has length a=6 and semi-minor axis has length b=13.

Step 2: Compute the distance of foci from the center (c)

For a hyperbola, the distance c of the foci from the center is given by the equation \(c=\sqrt{a^{2}+b^{2}}\). Substituting a=6 and b=13, \(c=\sqrt{(6^{2}+13^{2})} = \sqrt{205}\). Therefore, the foci are at (-sqrt{205}, 0) and (sqrt{205}, 0).

Step 3: Graphing the Hyperbola

The center of the hyperbola is at the origin. Draw the asymptotes as lines y= +/- (13/6)x. Using the semi-major axis a=6 and semi-minor axis b=13, draw a rectangle that fits within the asymptote lines. Draw the hyperbola with foci and vertices, which pass through the corners of the rectangle.

Key Concepts

Equation of a Hyperbola

The equation of a hyperbola can sometimes look deceptive, especially if you're not familiar with its standard form. A standard hyperbola equation can be distinguished by the subtraction between two squared terms set equal to 1, like \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \).
The orientation of the hyperbola—whether it opens horizontally or vertically—depends on which variable comes first. In the exercise, the equation is \( \frac{x^{2}}{36} - \frac{y^{2}}{169} = 1 \), where the \( x^2 \) term comes first, suggesting a horizontal hyperbola.Key components:

• The values under each squared term, \( a^2 \) and \( b^2 \), indicate the scales and orientations of the axes.
• The equation's numeral '1' implies the hyperbola is already simplified or in standard form so you don't need additional calculations to match it against the standard equation.

Foci of a Hyperbola

Understanding the foci of a hyperbola is key to grasping its unique properties. The foci points are integral because they help in shaping the hyperbola's curves. For a hyperbola, the foci are always located along the transverse axis (the real axis), which means they align with the direction the hyperbola opens. Finding the foci involves determining the distance from the center to each focus using the formula:\[c = \sqrt{a^2 + b^2}\]In our equation:

• \( a = 6 \) (since \( a^2 = 36 \)),
• \( b = 13 \) (since \( b^2 = 169 \)).

Substituting into the formula gives us:\[c = \sqrt{6^2 + 13^2} = \sqrt{205}\]Thus, the foci are located at \((-\sqrt{205}, 0)\) and \((\sqrt{205}, 0)\), focusing along the x-axis.
The foci represent the key points around which the hyperbola's two "arms" wrap, defining its extensive shape.

Graphing Hyperbolas

Graphing a hyperbola might feel complex, but breaking it into steps simplifies the task. The center point, vertex distances, foci, and asymptotes all serve as guides.Starting points:

• The center of this hyperbola is at \((0, 0)\), the origin.
• Draw the rectangle formed using \(2a = 12\) along the x-axis and \(2b = 26\) along the y-axis.Its sides run parallel to the axes.
• Draw lines from the corners of this rectangle, passing through the center, to represent the asymptotes. For this hyperbola: \( y = \pm \frac{13}{6}x \).

Now sketch hyperbola arcs that come near these lines, approaching but never touching them.With the rectangle, vertices (\((\pm 6, 0)\)), and asymptotes, it's straightforward to draw the hyperbola's arcs which loop around the rectangle and approach each line asymptotically.

Asymptotes of a Hyperbola

Asymptotes are lines that the branches of the hyperbola approach but never actually meet. They help in guiding the graphing of the hyperbola. For the standard horizontal hyperbola given by\( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the equations for the asymptotes are:\[ y = \pm \frac{b}{a}x \]In our example:

• The slopes of the asymptotes become \( \frac{13}{6} \), derived from \( b = 13 \) and \( a = 6 \).
• The equations are\( y = \pm \frac{13}{6}x \), showcasing symmetrical lines through the origin.

These asymptotes provide a clear framework for sketching the hyperbola, showing how steeply the branches open and the space they occupy as they stretch towards infinity on the graph.