Problem 66 - A Graphical Approach to College Algebra

Question

Find an equation for each hyperbola. \(y\) -intercept \((0,-2)\); center at origin; passing through \((2,3)\)

Step-by-Step Solution

Answer

\( \frac{y^2}{4} - \frac{5x^2}{16} = 1 \)

Step 1: Understand the Hyperbola

A hyperbola centered at the origin can be either of the form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), depending on the orientation. Since the \( y \)-intercept is \( (0, -2) \), it suggests that the hyperbola is vertical (\( y^2 \) comes first in the equation).

Step 2: Set Up the Hyperbola Equation

Using the form \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), substitute the \( y \)-intercept \( (0, -2) \). So, \( \frac{(-2)^2}{b^2} = 1 \), giving us \( \frac{4}{b^2} = 1 \). Therefore, \( b^2 = 4 \).

Step 3: Use the Passing Point

Now that \( b^2 = 4 \), substitute the given point \( (2, 3) \) into the equation \( \frac{y^2}{4} - \frac{x^2}{a^2} = 1 \). This becomes \( \frac{3^2}{4} - \frac{2^2}{a^2} = 1 \) or \( \frac{9}{4} - \frac{4}{a^2} = 1 \).

Step 4: Calculate \( a^2 \)

Solve \( \frac{9}{4} - \frac{4}{a^2} = 1 \) for \( a^2 \). Start by rearranging to \( \frac{9}{4} - 1 = \frac{4}{a^2} \). This simplifies to \( \frac{5}{4} = \frac{4}{a^2} \). Invert both sides to solve for \( a^2 \), obtaining \( a^2 = \frac{16}{5} \).

Step 5: Write the Equation

Substitute \( a^2 = \frac{16}{5} \) and \( b^2 = 4 \) back into the hyperbola equation \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \) to get \( \frac{y^2}{4} - \frac{x^2}{\frac{16}{5}} = 1 \). Simplify to \( \frac{y^2}{4} - \frac{5x^2}{16} = 1 \).

Key Concepts

Center at Origin

When examining hyperbolas, one of the most convenient starting points is to note the center's location. In this case, the hyperbola is centered at the

• origin, which means its coordinates are
  
• (0, 0).

This particular placement simplifies the hyperbola's equation, as no additional terms are needed to shift the graph horizontally or vertically on the coordinate plane. The standard form of a hyperbola's equation centered at the origin is either

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

The choice between these two forms hinges on the hyperbola's orientation—whether it opens horizontally or vertically. Understanding this is crucial, as it determines the structure of the equation and how the graph stretches.

Vertical Hyperbola

A hyperbola with a roof-like or vertical orientation is characterized by its focuses and vertices lying along the y-axis. In mathematical terms, this structure dictates that the

• \(y^2\) term comes first in the hyperbola's equation.

Given the exercise, the presence of a y-intercept suggests a vertical orientation, which makes our equation form: \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \). Observing the pattern where \( (0,-2) \) is a given point on the hyperbola immediately translates into the equation being:

• \( \frac{(-2)^2}{b^2} = 1 \).

Through this calculation, we find out that

• \( b^2 = 4 \).

Therefore, the hyperbola stretches vertically with a fixed measure of "spread" defined by \(b\). This width is paramount in determining the exact shape and form of the curve, once b is defined, our task is to ascertain the other aspect, the horizontal spread or "a" value.

Solving for a²

To determine the full equation of the hyperbola, it's essential to solve for \(a^2\) as well. The task becomes more interesting by using the point the hyperbola traverses, such as

• \( (2, 3) \).

Inserting this into the equation

• \( \frac{y^2}{4} - \frac{x^2}{a^2} = 1 \),

with

• \((x, y) = (2, 3)\),

we get:

• \( \frac{3^2}{4} - \frac{2^2}{a^2} = 1 \)

Simplifying this equation leads to:

• \( \frac{9}{4} - \frac{4}{a^2} = 1 \).

By rearranging terms, the equation simplifies to

• \( \frac{5}{4} = \frac{4}{a^2} \).

Cross-multiplying yields the solution:

• \( a^2 = \frac{16}{5} \).

Now with both \(a^2\) and \(b^2\) known, the precise equation of the vertical hyperbola is complete. Understanding the derivation and steps to find \(a^2\) is critical, allowing one to describe the curve fully and imagine its path across the coordinate plane.