Problem 10 - Precalculus: Mathematical for Calculus

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$9 x^{2}-4 y^{2}=36$$

Step-by-Step Solution

Answer

Vertices: (±2,0), Foci: (±√13,0), Asymptotes: y=±(3/2)x.

Step 1: Write in Standard Form

The given equation of the hyperbola is \(9x^2 - 4y^2 = 36\). To write this in standard form, divide by 36:\[ \frac{9x^2}{36} - \frac{4y^2}{36} = 1 \]This simplifies to \[ \frac{x^2}{4} - \frac{y^2}{9} = 1 \] The standard form is now \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where \(a^2 = 4\) and \(b^2 = 9\).

Step 2: Find the Vertices

For the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices are at \((\pm a, 0)\). Thus, \(a = \sqrt{4} = 2\). The vertices are \((2, 0)\) and \((-2, 0)\).

Step 3: Find the Foci

The foci of the hyperbola are given by \((\pm c, 0)\), where \(c^2 = a^2 + b^2\). Therefore, \[ c^2 = 4 + 9 = 13 \] \( c = \sqrt{13} \). So the foci are \((\sqrt{13}, 0)\) and \((-\sqrt{13}, 0)\).

Step 4: Find the Asymptotes

The equations of the asymptotes for the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are given by \[ y = \pm \frac{b}{a}x \] Substituting the values of \(a\) and \(b\), the asymptotes are \( y = \pm \frac{3}{2}x \).

Step 5: Sketch the Graph

To sketch the graph, first plot the center at the origin \((0, 0)\). Place the vertices at \((2, 0)\) and \((-2, 0)\). Plot the foci at \((\sqrt{13}, 0)\) and \((-\sqrt{13}, 0)\). Draw the asymptotes as lines through the origin with slopes \(\frac{3}{2}\) and \(-\frac{3}{2}\). Sketch the hyperbola opening left and right, approaching the asymptotes.

Key Concepts

Vertices of Hyperbola

In the world of hyperbolas, the vertices are key points that define the shape and orientation of the curve. These points are the closest to the center of the hyperbola. For hyperbolas represented by the equation \[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \],the vertices are located at \((\pm a, 0)\). This means that you can find the vertices by calculating \(a\), which is the square root of \(a^2\). In the exercise, with \(a^2 = 4\), we found \(a = 2\). Therefore, the vertices are \((2, 0)\) and \((-2, 0)\), lying along the x-axis. These vertices help us to understand the symmetry and the extent of the hyperbola along the horizontal axis.

Foci of Hyperbola

The foci of a hyperbola are two special points that have an important role in the definition of the curve. Unlike the vertices, these points do not lie on the hyperbola itself but rather indicate a unique property about the lengths of line segments connecting points on the hyperbola to the foci. For the standard hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \),the foci are positioned at \((\pm c, 0)\), where \(c\) is found using the equation \(c^2 = a^2 + b^2\). In our problem, with \(a^2 = 4\) and \(b^2 = 9\), we calculated \(c^2 = 13\), thus \(c = \sqrt{13}\). The foci, therefore, are \((\sqrt{13}, 0)\) and \((-\sqrt{13}, 0)\), which are crucial for understanding how the hyperbola stretches and aims to maintain its geometry relative to these focal points.

Asymptotes of Hyperbola

Asymptotes are lines that the hyperbola approaches but never actually touches or crosses. They provide a framework for sketching the hyperbola, giving you a sense of its general direction and curvature. For our hyperbola in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \),the equations of the asymptotes are given by \(y = \pm \frac{b}{a} x\). By substituting the values for \(a\) and \(b\),in our exercise, we found \(a = 2\) and \(b = 3\). Therefore, the asymptotes are \(y = \frac{3}{2}x\) and \(y = -\frac{3}{2}x\). These asymptotes define the slope of the diagonal lines that the branches of the hyperbola will approach. Understanding the role of asymptotes aids in accurately sketching the hyperbola's open and broad structure, typifying its behavior as it extends towards infinity.