Problem 33 - Thomas Calculus

Question

Exercises \(27-34\) give equations for hyperbolas. Put each equation in standard form and find the hyperbola's asymptotes. Then sketch the hyperbola. Include the asymptotes and foci in your sketch. $$ 8 y^{2}-2 x^{2}=16 $$

Step-by-Step Solution

Answer

Equation: \(\frac{y^2}{2} - \frac{x^2}{8} = 1\); Asymptotes: \(y = \pm \frac{1}{2}x\); Foci: \((0, \pm \sqrt{10})\).

Step 1: Rewrite Equation

Start by rewriting the given equation \(8y^{2} - 2x^{2} = 16\) in a simplified form. Divide every term by 16 to normalize the equation:\[ \frac{8y^2}{16} - \frac{2x^2}{16} = \frac{16}{16} \]This reduces to: \( \frac{y^2}{2} - \frac{x^2}{8} = 1 \).

Step 2: Identify Standard Hyperbola Form

Compare the simplified equation \( \frac{y^2}{2} - \frac{x^2}{8} = 1 \) with the standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Here, \(a^2 = 2\) and \(b^2 = 8\), therefore \(a = \sqrt{2}\) and \(b = 2\sqrt{2}\).

Step 3: Determine Asymptotes

For the hyperbola \( \frac{y^2}{2} - \frac{x^2}{8} = 1 \), the asymptotes are given by the lines:\[ y = \pm \frac{a}{b}x\]Here, \(a = \sqrt{2}\) and \(b = 2\sqrt{2}\). So, the asymptotes are \( y = \pm \frac{1}{2}x \).

Step 4: Identify Vertices and Foci

For the hyperbola \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the vertices are located at \((0, \pm a)\) and foci are at \((0, \pm c)\), where \(c = \sqrt{a^2 + b^2}\).Compute \(c\): \[ c = \sqrt{2 + 8} = \sqrt{10} \].Thus, the vertices are \((0, \pm \sqrt{2})\) and the foci are \((0, \pm \sqrt{10})\).

Step 5: Sketch the Hyperbola

Draw a coordinate system. Plot the vertices at \((0, \pm \sqrt{2})\) on the y-axis. Sketch the asymptotes \(y = \pm \frac{1}{2}x\). The hyperbola opens vertically, bounded by the asymptotes, and passes through the vertices. Mark the foci \((0, \pm \sqrt{10})\) on the y-axis inside the branches of the hyperbola.

Key Concepts

Standard Form of a Hyperbola

The standard form of a hyperbola is essential for easily understanding its overall shape and behavior. There are two common orientations of hyperbolas: horizontal and vertical. In a mathematical equation, the standard form looks like this for a vertically oriented hyperbola:

• \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \)

For the horizontal orientation, it is:

• \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \)

In our original problem, the hyperbola is vertical since the positive term is \(y^2\). To put the equation into standard form, divide all terms by the constant on the right side of the equation. This means we take the equation \( 8y^2 - 2x^2 = 16 \) and divide everything by 16 to get:

• \( \frac{y^2}{2} - \frac{x^2}{8} = 1 \)

This is now in standard form, with \(a^2 = 2 \) and \(b^2 = 8 \). It's important to remember that the standard form gives us direct information about the orientation and dimensions of the hyperbola.

Asymptotes of a Hyperbola

Asymptotes are the lines that a hyperbola approaches but never actually meets. They provide a guiding structure for sketching and understanding the hyperbola's open branches. For a hyperbola in standard form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the equations of the asymptotes are:

• \( y = \pm \frac{a}{b}x \)

In our exercise, we had values \( a = \sqrt{2} \) and \( b = 2\sqrt{2} \). Thus, the asymptotes are calculated as:

• \( y = \pm \frac{1}{2}x \)

These lines pass through the origin and have slopes of \( \pm \frac{1}{2} \). The asymptotes divide the plane into four sections and the branches of the hyperbola lie in the sections determined by these lines.Understanding these lines is crucial for creating an accurate sketch of the hyperbola.

Foci of a Hyperbola

The foci are another important feature of a hyperbola. They are two fixed points that lie along the axis of symmetry of the hyperbola. For an equation in the form \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \), the formula to find the coordinates of the foci is:

• \((0, \pm c )\) where \(c = \sqrt{a^2 + b^2}\)

In our example:

• \(a^2 = 2, \ b^2 = 8\)

Calculate \(c\) as:

• \(c = \sqrt{2 + 8} = \sqrt{10}\)

Thus, the foci of the hyperbola are located at \

• \((0, \pm \sqrt{10})\)

These points lie inside the branches of the hyperbola and are crucial for the definition as each point on the hyperbola maintains a constant difference in distance from these foci, highlighting the unique behavior of the curve.