Problem 31 - Precalculus: Functions and Graphs

Question

Find an equation for the hyperbola that has its center at the origin and satisfies the given conditions. Foci \(F(0, \pm 10), \quad\) asymptotes \(y=\pm \frac{1}{3} x\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \(\frac{y^2}{10} - \frac{x^2}{90} = 1\).

Step 1: Understand the Standard Equation Form for a Hyperbola

The standard form of a hyperbola centered at the origin can be either \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) or \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\). Since the foci are on the y-axis \((0, \pm 10)\), the equation of our hyperbola will be \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\).

Step 2: Use the Asymptotes to Determine the Relationship Between a and b

The equations of the asymptotes for the hyperbola \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\) are \(y = \pm \frac{b}{a}x\). Given the asymptotes are \(y = \pm \frac{1}{3}x\), we have \(\frac{b}{a} = \frac{1}{3}\). Thus, \(b = \frac{a}{3}\).

Step 3: Use the Foci to Determine b

The distance from the center of the hyperbola to the foci is given by \(c = 10\). For a vertical hyperbola, \(c = \sqrt{a^2 + b^2}\). Substitute the value of \(b\) from Step 2: \(c = \sqrt{a^2 + \left(\frac{a}{3}\right)^2}\). Substitute \(c = 10\) and solve: \(10 = \sqrt{a^2 + \frac{a^2}{9}}\).

Step 4: Solve for a and b

Solve the equation from Step 3: \(10 = \sqrt{a^2 + \frac{a^2}{9}}\). Square both sides: \(100 = a^2 + \frac{a^2}{9}\). Write as a common denominator: \(100 = \frac{9a^2 + a^2}{9}\), so \(900 = 10a^2\). Solve for \(a^2\): \(a^2 = 90\). Since \(b = \frac{a}{3}\), we have \(b^2 = \left(\frac{a}{3}\right)^2 = \frac{a^2}{9} = 10\).

Step 5: Write the Equation of the Hyperbola

With \(a^2 = 90\) and \(b^2 = 10\), substitute back into the hyperbola equation: \(\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1\). Therefore, the equation of the hyperbola is \(\frac{y^2}{10} - \frac{x^2}{90} = 1\).

Key Concepts

Conic Sections

Conic sections are the curves obtained by intersecting a plane with a double-napped cone. They include circles, ellipses, parabolas, and hyperbolas. Each conic section has unique properties and equations that define their shapes. Hyperbolas, one of these conic sections, are characterized by their open curves. Unlike other conics, such as circles and ellipses, where you have closed loops, hyperbolas open outward, either vertically or horizontally.

Hyperbolas are quite unique because they consist of two separate curves, called branches. These branches mirror each other and are symmetric around the center of the hyperbola. This centered symmetry is essential in defining the structure of hyperbolas. For hyperbolas, knowing the orientation is crucial, as it determines the form of their equation. In our case, with the center at the origin, and foci along the y-axis, results in a vertically oriented hyperbola.

Asymptotes

Asymptotes of a hyperbola are straight lines that the curve approaches but never touches or crosses. They guide the shape of the hyperbola's branches. The asymptotes intersect at the center (origin) of the hyperbola. For our hyperbola, the asymptotes are defined as:

• \( y = \frac{1}{3}x \)
• \( y = -\frac{1}{3}x \)

These lines tell us how the hyperbola forms and the direction it opens. By comparing these asymptotes with the standard form's asymptote equation \( y = \pm \frac{b}{a}x \), we can find the relationship between \(a\) and \(b\). This connection reveals how steep or wide the branches of the hyperbola will be as they stretch into infinity.

Foci

The foci of a hyperbola are two fixed points located inside each branch of the hyperbola. These points show how 'stretched' the hyperbola is. They are essential to its geometric properties. The distance from the center to each focus indicates how broad or narrow the hyperbola will appear.

In the given exercise, the foci are at \((0, \pm 10)\). This confirms that our hyperbola is vertically oriented, emphasizing that its branches open up and down along the y-axis. The concept of foci plays a crucial role in determining the hyperbola's equation. The distance to the foci, represented as \(c\), integrates into the formula \(c = \sqrt{a^2 + b^2}\) to help find the equation's correct values.

Standard Equation of Hyperbola

The standard equation for a hyperbola depends on its orientation and can come in two forms. For hyperbolas centered at the origin, these two forms are:

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

In our problem, we use the vertical equation form because the foci are along the y-axis. Using the conditions provided such as foci and asymptotes, we solve for \(a^2\) and \(b^2\) as part of steps to form the hyperbola's equation.

Finally, using \(a^2 = 90\) and \(b^2 = 10\), the exact equation of this hyperbola becomes \( \frac{y^2}{10} - \frac{x^2}{90} = 1 \), capturing its unique geometric structure.