Problem 16 - Algebra 2

Question

Find the coordinates of the vertices and foci and the equations of the asymptotes for the hyperbola with the given equation. Then graph the hyperbola. $$ \frac{x^{2}}{81}-\frac{y^{2}}{49}=1 $$

Step-by-Step Solution

Answer

Vertices: (9, 0), (-9, 0); Foci: (√130, 0), (-√130, 0); Asymptotes: y=±7/9x.

Step 1: Identify the Standard Form of the Hyperbola

The given hyperbola equation is \( \frac{x^{2}}{81} - \frac{y^{2}}{49} = 1 \). This matches the standard form of a hyperbola centered at the origin, which is \( \frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1 \). From this, we can identify \( a^{2} = 81 \) and \( b^{2} = 49 \). Thus, \( a = 9 \) and \( b = 7 \). Since the \( x^2 \) term is positive, the transverse axis is horizontal.

Step 2: Find the Vertices

For a hyperbola with a horizontal transverse axis centered at the origin, the vertices are located at \((\pm a, 0)\). Substituting \(a = 9\), the vertices are at \((9, 0)\) and \((-9, 0)\).

Step 3: Find the Foci

The foci of a hyperbola with a horizontal transverse axis are found at \((\pm c, 0)\), where \(c\) is calculated using \( c = \sqrt{a^2 + b^2} \). Here, \( c = \sqrt{81 + 49} = \sqrt{130} \). Thus, the foci are at \((\sqrt{130}, 0)\) and \((- \sqrt{130}, 0)\).

Step 4: Determine the Asymptotes

For a hyperbola of the form \(\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1\), the equations of the asymptotes are \(y = \pm \frac{b}{a}x\). Substituting \(a = 9\) and \(b = 7\), the equations of the asymptotes are \(y = \pm \frac{7}{9}x\).

Step 5: Graph the Hyperbola

The graph of the hyperbola will be centered at the origin with vertices at \((9, 0)\) and \((-9, 0)\), foci at \((\sqrt{130}, 0)\) and \((- \sqrt{130}, 0)\), and asymptotes given by \(y = \frac{7}{9}x\) and \(y = -\frac{7}{9}x\). The branches of the hyperbola open left and right along the horizontal axis. Plot these key features to draw the hyperbola.

Key Concepts

Vertices

In a hyperbola, the vertices mark the points where the hyperbola is closest to its center. Identifying these crucial points helps to understand the hyperbola's orientation and spread. For the hyperbola given by the equation \[ \frac{x^{2}}{81} - \frac{y^{2}}{49} = 1, \]we first determine the values of \(a\) and \(b\) from the denominators. Here, \(a^2 = 81\) gives us \(a = 9\). The equation describes a hyperbola with a horizontal transverse axis, as the \(x^2\) term is positive. Therefore, the vertices are located at \((\pm a, 0)\). For this hyperbola, that means the vertices are \((9, 0)\) and \((-9, 0)\).

• Vertices: These are the endpoints of the transverse axis of the hyperbola.
• Coordinate Calculation: Use \((\pm a, 0)\) for a horizontal transverse axis.
• This hyperbola's vertices: \((9, 0)\) and \((-9, 0)\).

Foci

The foci of a hyperbola are another pair of special points. They are located further from the center than the vertices. The foci determine the shape of the hyperbola and help in measuring the eccentricity. For the same hyperbola equation, \[ \frac{x^{2}}{81} - \frac{y^{2}}{49} = 1, \] we calculate the distance of the foci from the center by considering the equation \[ c = \sqrt{a^2 + b^2} \] here, \(a^2 = 81\) and \(b^2 = 49\). Thus,\[ c = \sqrt{81 + 49} = \sqrt{130}. \]The foci are positioned at \((\pm c, 0)\) for a hyperbola with a horizontal axis, giving \((\sqrt{130}, 0)\) and \((- \sqrt{130}, 0)\).

• Foci Role: Important for understanding the hyperbola's stretching.
• Distance Calculation: Use \( c = \sqrt{a^2 + b^2} \) to find \( c \).
• This hyperbola's foci: \((\sqrt{130}, 0)\) and \((- \sqrt{130}, 0)\).

Asymptotes

Asymptotes of a hyperbola act as invisible boundaries that the branches of the hyperbola approach but never touch. They give insight into the hyperbola's widening. For our hyperbola, given by \[ \frac{x^{2}}{81} - \frac{y^{2}}{49} = 1, \]the equations of the asymptotes are represented by \[ y = \pm \frac{b}{a}x. \]Here, substituting \(a = 9\) and \(b = 7\), we find the asymptote equations as\[ y = \pm \frac{7}{9}x. \]

• Role of Asymptotes: Guide the paths of the hyperbola branches.
• Equation Format: Generally, use \( y = \pm \frac{b}{a}x \).
• Asymptotes here: \( y = \frac{7}{9}x \) and \( y = -\frac{7}{9}x \).

Standard Form

The standard form of a hyperbola provides critical information at a glance, outlining its orientation and dimensions. The general standard form of a hyperbola is\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]for a horizontally oriented hyperbola. By determining the denominators, one can ascertain \(a^2\) and \(b^2\) and then those help locate vertices and foci. For our equation,\[ \frac{x^{2}}{81} - \frac{y^{2}}{49} = 1, \]we have \(a^2 = 81\) and \(b^2 = 49\), thus \(a = 9\) and \(b = 7\). The equation shows it is horizontally oriented because the \(x\) term is positive.

• Form Significance: Identifies axis orientation (horizontal/vertical)
• Components: Compare with \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) to find \(a\) and \(b\).
• This hyperbola's parameters: \(a = 9, b = 7\).

Graphing Hyperbolas

Graphing hyperbolas involves plotting several key features: the center, vertices, foci, and asymptotes. This comprehensive approach ensures clarity and accuracy. Start by marking the center of the hyperbola at the origin since it's in standard form and centered there. Next, plot the vertices at\((9, 0)\) and \((-9, 0)\).Foci, located at \((\sqrt{130}, 0)\) and \((- \sqrt{130}, 0)\), give insight into the curve direction. Asymptotes, with the equations\[ y = \frac{7}{9}x \] and \[ y = -\frac{7}{9}x, \]act as the guiding lines for the hyperbola's shape. Draw the hyperbola branches to touch these vertices and gradually open outwards, always approaching but never crossing the asymptotes.

• Steps for Graphing: Mark the center and plot vertices first.
• Include Asymptotes: Draw as straight lines through the center at the slopes given.
• This hyperbola: Center at origin, branches open along the horizontal.