Problem 7 - Precalculus

Question

Graph the hyperbolas. In each case in which the hyperbola is nondegenerate, specify the following: vertices, foci, lengths of transverse and conjugate axes, eccentricity, and equations of the asymptotes. also specify The centers. $$2 y^{2}-3 x^{2}=1$$

Step-by-Step Solution

Answer

Vertices: (0, ±√2/2), Foci: (0, ±√5/6), Center: (0, 0), Transverse axis length: √2, Conjugate axis length: 2/√3, Eccentricity: √10/3, Asymptotes: y = ±√6/2 x.

Step 1: Rewrite the Equation in Standard Form

The given equation is \(2y^2 - 3x^2 = 1\). Divide the whole equation by 1 to express it in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). This gives us \(\frac{y^2}{\frac{1}{2}} - \frac{x^2}{\frac{1}{3}} = 1\). Thus, \(a^2 = \frac{1}{2}\) and \(b^2 = \frac{1}{3}\).

Step 2: Identify the Center

The given equation \(2y^2 - 3x^2 = 1\) is in the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) with no terms added or subtracted from \(x\) or \(y\), which indicates that the center is at the origin \((0, 0)\).

Step 3: Determine the Vertices

For a hyperbola of the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the vertices are located at \((0, \pm a)\). Since \(a^2 = \frac{1}{2}\), it follows that \(a = \sqrt{\frac{1}{2}} = \frac{\sqrt{2}}{2}\). Thus, the vertices are \((0, \pm \frac{\sqrt{2}}{2})\).

Step 4: Calculate Foci

The foci of the hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) are located at \((0, \pm c)\) where \(c = \sqrt{a^2 + b^2}\). Thus, \(c^2 = \frac{1}{2} + \frac{1}{3} = \frac{3}{6} + \frac{2}{6} = \frac{5}{6}\). Therefore, \(c = \sqrt{\frac{5}{6}}\) making the foci at \((0, \pm \sqrt{\frac{5}{6}})\).

Step 5: Find Lengths of Transverse and Conjugate Axes

The length of the transverse axis for \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) is \(2a\) and conjugate axis is \(2b\). Thus, transverse axis length is \(2 \times \frac{\sqrt{2}}{2} = \sqrt{2}\) and conjugate axis length is \(2 \times \sqrt{\frac{1}{3}} = \frac{2}{\sqrt{3}}\).

Step 6: Calculate the Eccentricity

Eccentricity \(e\) of a hyperbola is given by \(e = \frac{c}{a}\). From previous steps, \(c = \sqrt{\frac{5}{6}}\) and \(a = \frac{\sqrt{2}}{2}\). Thus, \(e = \frac{\sqrt{\frac{5}{6}}}{\frac{\sqrt{2}}{2}} = \frac{2\sqrt{\frac{5}{6}}}{\sqrt{2}}\). Simplifying gives \(e = \sqrt{\frac{10}{3}}\).

Step 7: Find Equations of Asymptotes

The asymptotes for \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) are given by the equations \(y = \pm \frac{a}{b}x\). Therefore, \(y = \pm \frac{\frac{\sqrt{2}}{2}}{\sqrt{\frac{1}{3}}}x = \pm \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{1} x = \pm \frac{\sqrt{6}}{2}x\).

Key Concepts

Hyperbola Vertices

Hyperbola vertices are specific points where the hyperbola reaches its closest distance from the center along the transverse axis. These are crucial in defining the size and shape of the hyperbola.

• For a standard hyperbola equation \[\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\], the vertices are located along the y-axis (for vertical opening), specifically at \((0, \pm a)\).
• In our given equation \(2y^2 - 3x^2 = 1\), we wrote it as \[\frac{y^2}{\frac{1}{2}} - \frac{x^2}{\frac{1}{3}} = 1\]. Here, \(a^2 = \frac{1}{2}\), meaning \(a = \frac{\sqrt{2}}{2}\). Therefore, the vertices are at \((0, \pm \frac{\sqrt{2}}{2})\).

These vertices are fundamental to sketching the graph of a hyperbola, serving as essential markers on the curve.

Hyperbola Foci

The foci of a hyperbola are critical in understanding its geometric properties. They are positioned further out than the vertices, and each branch of the hyperbola approaches these points.

• The coordinates of the foci for the equation \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) are determined as \((0, \pm c)\), where \(c = \sqrt{a^2 + b^2}\).
• For our hyperbola, \[c^2 = \frac{1}{2} + \frac{1}{3} = \frac{5}{6}\]. Hence, \(c = \sqrt{\frac{5}{6}}\), leading to the foci positioned at \((0, \pm \sqrt{\frac{5}{6}})\).

These points are important as they define the hyperbola's eccentricity and help in understanding its skew from a perfect circular shape.

Hyperbola Asymptotes

Asymptotes play a crucial role in sketching hyperbolas as they indicate direction for the branches of the hyperbola, acting like guidelines. Asymptotes are straight lines that the hyperbola approaches but never touches.

• For hyperbolas given by \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the equations of the asymptotes are \(y = \pm \frac{a}{b}x\).
• In our case, \(a = \frac{\sqrt{2}}{2}\) and \(b = \sqrt{\frac{1}{3}}\). Calculating gives: \[y = \pm \frac{\frac{\sqrt{2}}{2}}{\sqrt{\frac{1}{3}}}x = \pm \frac{\sqrt{6}}{2}x\].

These lines provide an invaluable reference while sketching the hyperbola and understanding its orientation.

Transverse Axis

The transverse axis is an important concept in hyperbolas as it relates to the orientation of the hyperbola's branches and the positioning of specific attributes like vertices and foci.

• It is the axis along which the vertices are located. For \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the transverse axis is vertical, coinciding with the y-axis.
• The length of the transverse axis is given by \(2a\). For this problem, \(a = \frac{\sqrt{2}}{2}\), making the transverse axis \(\sqrt{2}\).

This axis reflects the stretch of the hyperbola in the direction of vertex alignment, giving insight into the hyperbola's "height."

Conjugate Axis

The conjugate axis is perpendicular to the transverse axis and defines the span of a hyperbola away from its vertices.

• For the hyperbola \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\), the conjugate axis is horizontal, aligning along the x-axis.
• Its length is determined by \(2b\), where \(b = \sqrt{\frac{1}{3}}\). Thus, the conjugate axis is \(\frac{2}{\sqrt{3}}\).

Understanding the conjugate axis helps to comprehend the extent of the hyperbola's width and is also crucial in locating the asymptotes.

Eccentricity

Eccentricity quantifies how much a conic section deviates from being circular. For hyperbolas, it determines the openess of the curves.

• Eccentricity (\(e\)) for hyperbolas is calculated via \(e = \frac{c}{a}\), where \(c\) is the distance to foci and \(a\) is distance to vertices.
• In this hyperbola, where \(c = \sqrt{\frac{5}{6}}\) and \(a = \frac{\sqrt{2}}{2}\), the eccentricity is: \[e = \frac{\sqrt{\frac{5}{6}}}{\frac{\sqrt{2}}{2}} = \sqrt{\frac{10}{3}}\].

Higher eccentricity values mean the hyperbola is more elongated, providing insights into how stretched the hyperbola is away from being a circle.