Problem 18 - Precalculus Mathematics 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: \(y = \pm \frac{1}{\sqrt{2}}x\).

Step 1: Identify the Standard Form

To find the vertices, foci, and asymptotes of the hyperbola \(x^2 - 2y^2 = 3\), we first need to rewrite it in its standard form. The standard form of a hyperbola centered at the origin is \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\). Divide the original equation by 3 to obtain \(\frac{x^2}{3} - \frac{y^2}{1.5} = 1\). Here, \(a^2 = 3\) and \(b^2 = 1.5\).

Step 2: Calculate Vertices

The vertices of a hyperbola in standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are located at \((\pm a, 0)\). Therefore, compute \(a = \sqrt{3}\), which gives the vertices as \((\sqrt{3}, 0)\) and \((-\sqrt{3}, 0)\).

Step 3: Calculate Foci

The foci of a hyperbola are located at \((\pm c, 0)\), where \(c = \sqrt{a^2 + b^2}\). Calculate \(c = \sqrt{3 + 1.5} = \sqrt{4.5}\). Therefore, the foci are at \((\sqrt{4.5}, 0)\) and \((-\sqrt{4.5}, 0)\).

Step 4: Determine Asymptotes

The asymptotes of a hyperbola in standard form \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) are given by the equations \(y = \pm \frac{b}{a}x\). Here, \(\frac{b}{a} = \frac{\sqrt{1.5}}{\sqrt{3}} = \frac{1}{\sqrt{2}}\). Thus, the asymptotes are \(y = \pm \frac{1}{\sqrt{2}}x\).

Step 5: Sketch the Graph

Plot the vertices \((\pm \sqrt{3}, 0)\) and the asymptotes \(y = \pm \frac{1}{\sqrt{2}}x\) on the coordinate plane. Draw the hyperbola, opening left and right, centered at the origin, and approaching the asymptotes as it extends away from the center.

Key Concepts

Vertices of a Hyperbola

When dealing with the geometry of a hyperbola, one of the key features you will calculate is the vertices. Vertices are points on the hyperbola that are closest to the center. They represent the turning points from which each branch of the hyperbola curves away.

For the hyperbola equation in standard form, given by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the vertices can be found at position \((\pm a, 0)\) if the hyperbola opens horizontally. Here, \(a\) is the semi-major axis, and solving \(a = \sqrt{3} \) from our equation gives the vertex positions as \((\sqrt{3}, 0)\) and \(( -\sqrt{3}, 0)\).

• The vertices are significant because they determine the width of the box around which the hyperbola curves.
• Pay attention to these points as they give fundamental direction to the opening of the hyperbola.

Foci of a Hyperbola

Foci are crucial in defining the shape and orientation of a hyperbola. A hyperbola's foci lie on the same axis as the vertices but are located further away from the center.

The foci of a hyperbola described by \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are found at \((\pm c, 0)\), where \(c\) is determined by \(c = \sqrt{a^2 + b^2}\). In our example, substituting the values gives \(c = \sqrt{4.5} \approx 2.12\), placing the foci at approximately \((2.12, 0)\) and \((-2.12, 0)\).

• The foci are a bit further than the vertices, giving each hyperbola branch its distinct curvature.
• Position of the foci indicates how 'stretched' the hyperbola is.

Asymptotes of a Hyperbola

Asymptotes in hyperbolas act as guide lines that the curve approaches but never intersects. They dictate the general direction that each branch of the hyperbola takes.

For the hyperbola in standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the asymptotes are defined by the equations \(y = \pm \frac{b}{a}x \). For our problem, this ratio is \( \frac{1}{\sqrt{2}} \), resulting in the asymptotes \(y = \pm \frac{1}{\sqrt{2}} x\).

• The asymptotes cross at the hyperbola's center and form a diagonal 'X' that frames and guides the hyperbola's shape.
• Understanding where the asymptotes are can help you sketch a more accurate hyperbola.