Problem 14 - Precalculus: Mathematical for Calculus

Question

Find the vertices, foci, and asymptotes of the hyperbola, and sketch its graph. $$x^{2}-2 y^{2}=3$$

Step-by-Step Solution

Answer

Vertices: \((\pm \sqrt{3}, 0)\), Foci: \((\pm \sqrt{4.5}, 0)\), Asymptotes: \(y = \pm \frac{\sqrt{2}}{2} x\).

Step 1: Write the Standard Form of the Hyperbola

A hyperbola's standard form is either \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) or \( \frac{(y-k)^2}{b^2} - \frac{(x-h)^2}{a^2} = 1 \). Reorganize the given equation, \( x^2 - 2y^2 = 3 \), into this form.Divide through by 3 to obtain: \( \frac{x^2}{3} - \frac{y^2}{1.5} = 1 \).Thus, the hyperbola's standard form is \( \frac{x^2}{3} - \frac{y^2}{1.5} = 1 \).

Step 2: Identify the Hyperbola Components

From the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we see that \( a^2 = 3 \) and \( b^2 = 1.5 \).So, \( a = \sqrt{3} \) and \( b = \sqrt{1.5} \). The center \((h, k)\) is at \( (0, 0) \).

Step 3: Find the Vertices

For the hyperbola, the transverse axis is along the x-axis because \( \frac{x^2}{a^2} \) term is positive.The vertices are at \( (\pm a, 0) \).Substituting \( a = \sqrt{3} \), the vertices are \( (\sqrt{3}, 0) \) and \( (-\sqrt{3}, 0) \).

Step 4: Find the Foci

The distance of the foci from the center along the x-axis for this hyperbola is \( c \), where \( c = \sqrt{a^2 + b^2} \).Calculate: \( c = \sqrt{3 + 1.5} = \sqrt{4.5} \).Thus, the foci are at \( (\pm \sqrt{4.5}, 0) \).

Step 5: 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 \( a = \sqrt{3} \) and \( b = \sqrt{1.5} \), we have the asymptotes:\( y = \pm \frac{\sqrt{1.5}}{\sqrt{3}} x = \pm \frac{1}{\sqrt{2}} x = \pm \frac{\sqrt{2}}{2} x \).

Step 6: Sketch the Hyperbola

To sketch the hyperbola, plot the center at \((0, 0)\), vertices \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\), and the foci at \((\sqrt{4.5}, 0)\) and \((-\sqrt{4.5}, 0)\).Draw the asymptotes, \( y = \pm \frac{\sqrt{2}}{2} x \), as dashed lines. Sketch the hyperbola opening sideways (along the x-axis) through its vertices.

Key Concepts

Vertices of Hyperbola

The vertices of a hyperbola are crucial points that help in defining the shape and orientation of the graph. In the context of a hyperbola whose equation is given in the standard form of \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the vertices lie along the transverse axis. This axis runs through the center of the hyperbola and is parallel to either the x or y-axis, depending on the term with the leading coefficient. For our particular hyperbola described by \(\frac{x^2}{3} - \frac{y^2}{1.5} = 1\), the transverse axis is aligned with the x-axis because the x-term has a positive coefficient.

To find the vertices, we focus on the transverse axis and the value of \(a\), which is the distance from the center to each vertex. Here, \(a = \sqrt{3}\), and the center of our hyperbola is at the origin \((0, 0)\). The vertices thus are located at \((a, 0)\) and \((-a, 0)\), resulting in the vertices being \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\). These points represent the extreme points along the major axis of the hyperbola.

The position of vertices ultimately determines the spread and direction of the hyperbola. For a hyperbola opening left and right along the x-axis, like in this case, the vertices are horizontally aligned at these same points.

Foci of Hyperbola

Foci are a defining feature of hyperbolas, located along the transverse axis, further out than the vertices. They are critical to the geometric definition of a hyperbola. The hyperbola stretches outward such that the absolute difference of the distances from any point on its branches to the two foci is constant. In the equation \(\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1\), the distance \(c\) from the center to each focus is calculated using \(c = \sqrt{a^2 + b^2}\).

For our specific hyperbola, where \(a = \sqrt{3}\) and \(b = \sqrt{1.5}\), we find \(c\) by calculating \(\sqrt{3 + 1.5} = \sqrt{4.5}\). This gives us the location of the foci at \((\pm \sqrt{4.5}, 0)\).

These points are more spread apart than the vertices and provide key insights into the hyperbola's width and eccentricity. As the foci get further apart, the branches of the hyperbola become less steep, indicating greater separation.

Asymptotes of Hyperbola

Asymptotes are lines that a hyperbola approaches but never actually meets. For a hyperbola, these diagonally running lines guide the curve's end behavior as it extends infinitely. In our hyperbola's standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\), the equations of the asymptotes are derived as \(y = \pm \frac{b}{a}x\).

Here, using our values \(b = \sqrt{1.5}\) and \(a = \sqrt{3}\), the asymptotes are calculated to be \(y = \pm \frac{\sqrt{1.5}}{\sqrt{3}}x\). Simplifying further, these equations become \(y = \pm \frac{1}{\sqrt{2}}x = \pm \frac{\sqrt{2}}{2}x\).

These asymptotes help us visualize where the hyperbola's branches will head as they stretch out from the center, increasing the accuracy when sketching the graph. They intersect at the hyperbola's center, providing a crosshair for symmetry and balance in its drawing.