Problem 32 - Precalculus with Calculus Previews

Question

In Problems \(21-44,\) find an equation of the hyperbola that satisfies the given conditions. Foci \((0,\pm 3),\) asymptotes \(y=\pm \frac{3}{2} x\)

Step-by-Step Solution

Answer

The equation of the hyperbola is \(\frac{13y^2}{36} - \frac{13x^2}{27} = 1\).

Step 1: Determine the Orientation of the Hyperbola

Given the foci at \((0, \pm 3)\), we can see that they are on the y-axis. This tells us the hyperbola is vertically oriented.

Step 2: Identify Values from Foci

For vertical hyperbolas, the foci coordinates are of the form \((0, \pm c)\). Here, \(c = 3\).Thus, the center of the hyperbola is at the origin \((0, 0)\).

Step 3: Identify Values from Asymptotes

The equations of the asymptotes for a vertical hyperbola centered at the origin are of the form \(y = \pm \frac{b}{a} x\). Given the asymptotes \(y = \pm \frac{3}{2} x\), we can deduce that \(\frac{b}{a} = \frac{3}{2}\).

Step 4: Relate Parameters with Horizontal and Vertical Radii

The relation between parameters is given by the equation \(a^2 + b^2 = c^2\). Here, \(c^2 = 9\).

Step 5: Solve for \(a\) and \(b\)

We know from asymptotes that \(\frac{b}{a} = \frac{3}{2}\), which implies \(b = \frac{3}{2}a\). Substitute \(b\) in the relation to solve for \(a\):\[a^2 + \left(\frac{3}{2}a\right)^2 = 9\]Simplifying gives:\[a^2 + \frac{9}{4}a^2 = 9\]\[\frac{13}{4}a^2 = 9\]\[a^2 = \frac{36}{13}\]Thus, \(b^2 = \left(\frac{3}{2}a\right)^2 = \frac{27}{13}\).

Step 6: Write Down the Equation of the Hyperbola

For a vertical hyperbola centered at the origin, the equation is of the form \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Substitute \(a^2\) and \(b^2\): \[\frac{y^2}{\frac{36}{13}} - \frac{x^2}{\frac{27}{13}} = 1\]Simplify to get:\[\frac{13y^2}{36} - \frac{13x^2}{27} = 1\]This is the equation of the hyperbola.

Key Concepts

Foci

In the context of hyperbolas, the foci are two special points located on the principal axis of the hyperbola. These points have the property that the difference in distances from any point on the hyperbola to each focus is constant.
For a hyperbola that is oriented vertically, the foci are positioned along the y-axis.

In this specific exercise, the foci are given as

• (0, 3)
• (0, -3)

This tells us that the hyperbola is vertically oriented, with its center at the origin,

• (0, 0).

The distance from the center to each focus, denoted as "c", is 3 in this case.

This information is crucial in determining other properties of the hyperbola, like its shape and equation.

Asymptotes

Asymptotes are lines that a curve approaches as it heads towards infinity.
For hyperbolas, the asymptotes give a guideline for the hyperbola's steepness and general direction.

In the given exercise, we have the asymptotes

• y = \(\frac{3}{2}x\)
• y = -\(\frac{3}{2}x\)

These equations suggest that the orientation of our hyperbola's branches is vertical, converging towards these lines as \(|y|\rightarrow \infty\).

Knowing the slope of the asymptotes helps us relate the semi-major axis "a" and semi-minor axis "b" of the hyperbola. For vertical hyperbolas, the slopes are given by \(\frac{b}{a}\). Here, we find that:

• \(\frac{b}{a} = \frac{3}{2}\)

This ratio is essential in solving for "a" and "b" using the relationship derived from the asymptotes.

Equation of a hyperbola

The equation of a hyperbola allows us to represent it algebraically. For a hyperbola centered at the origin, the standard form depends on its orientation.
A vertically oriented hyperbola takes the form:

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

In our exercise, finding the equation required deriving "a" and "b" using the foci and asymptotes. By utilizing the relationship \(a^2 + b^2 = c^2\), with \(c = 3\), we substitute \(b\) as \(\frac{3}{2}a\), known from the asymptotic slope:

• \(a^2 + \left(\frac{3}{2}a\right)^2 = 9\)
• Simplifying: \(a^2 = \frac{36}{13}\)
• Therefore, \(b^2 = \left(\frac{3}{2}a\right)^2 = \frac{27}{13}\)

Substituting into the standard form:

• \(\frac{y^2}{\frac{36}{13}} - \frac{x^2}{\frac{27}{13}} = 1\)

Ultimately, it simplifies to:

• \(\frac{13y^2}{36} - \frac{13x^2}{27} = 1\)

This complete equation captures the unique characteristics and orientation of our hyperbola.