Problem 43 - Precalculus: Mathematics for Calculus

Question

Find an equation for the hyperbola that satisfies the given conditions. Vertices: \((0, \pm 6),\) hyperbola passes through \((-5,9)\)

Step-by-Step Solution

Answer

\(\frac{y^2}{36} - \frac{x^2}{20} = 1\)

Step 1: Identify the Type of Hyperbola

Given the vertices (0, \pm 6), the transverse axis is vertical, and the center is at the origin (0,0). This indicates a vertical hyperbola of the form\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \] where \(a\) represents the distance from the center to the vertices.

Step 2: Use the Vertices to Find \(a^2\)

The distance from the center to each vertex is 6 units, so \(a = 6\). Therefore, \(a^2 = 36\). The initial equation of the hyperbola becomes:\[ \frac{y^2}{36} - \frac{x^2}{b^2} = 1 \]

Step 3: Substitute the Point (-5, 9)

Substitute \(x = -5\) and \(y = 9\) into the equation to find \(b^2\):\[ \frac{9^2}{36} - \frac{(-5)^2}{b^2} = 1 \]Simplify to obtain:\[ \frac{81}{36} - \frac{25}{b^2} = 1 \]

Step 4: Solve for \(b^2\)

Simplify the left side of the equation:\[ \frac{9}{4} - \frac{25}{b^2} = 1 \]Move \( \frac{25}{b^2} \) to the right side and 1 to the left:\[ \frac{9}{4} - 1 = \frac{25}{b^2} \]\[ \frac{5}{4} = \frac{25}{b^2} \]Multiply both sides by \(b^2\) and solve for \(b^2\):\[ 5b^2 = 100 \]\[ b^2 = 20 \]

Step 5: Write the Equation of the Hyperbola

Substituting \(b^2 = 20\) back into the hyperbola equation, we have:\[ \frac{y^2}{36} - \frac{x^2}{20} = 1 \]

Step 6: Conclusion

The equation of the hyperbola with vertices at (0, \pm 6) and passing through (-5,9) is:\[ \frac{y^2}{36} - \frac{x^2}{20} = 1 \]

Key Concepts

Equation of Hyperbola

The equation of a hyperbola is crucial as it defines the curve and its properties. Each hyperbola will have its unique equation based on its orientation (whether it's horizontal or vertical) and its specific characteristics. In the typical Cartesian coordinate system, a hyperbola centered at the origin can either open vertically or horizontally, and this orientation determines the structure of the equation. For a vertical hyperbola, the equation is of the form:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]Here:

• \(a^2\) and \(b^2\) are the squares of the distances from the center to the vertices and co-vertices, respectively.
• The terms involving \(y\) and \(x\) reflect how the hyperbola opens and stretches along these axes.

To determine the specific equation for a given set of conditions, one must know these critical values and the hyperbola's orientation, as demonstrated in the above steps.

Vertices of Hyperbola

The vertices of a hyperbola are key points on the curve that represent the closest (or farthest in case of a standard definite measurement) points to the center. In a vertical hyperbola, these vertices lie along the \(y\)-axis, whereas in a horizontal hyperbola, they are along the \(x\)-axis. In our exercise, the vertices are located at \((0, \pm 6)\). This provides important information:

• The distance from the center to each vertex (denoted as \(a\)) is seen to be 6.
• This informs us that \(a^2 = 36\) as \(a\) is squared in the hyperbola equation.

Understanding the vertices allows us to establish part of the standard form equation, giving structure to our hyperbola's mathematical description.

Center of Hyperbola

The center of a hyperbola is the midpoint between its vertices and serves as the origin for defining all other significant points of the hyperbola, such as foci and co-vertices. When a hyperbola is centered at the origin \((0,0)\), the equation takes a simplified standard form, making calculations and interpretations straightforward. Knowing the center:

• Helps determine the directional orientation of the hyperbola.
• Assists in setting initial values for positioning the hyperbola on a coordinate plane.

In practical problems, establishing the center is the starting point for constructing the rest of the hyperbola's characteristics, making it essential to properly solve real-world geometric problems.

Transverse Axis

The transverse axis of a hyperbola is the axis along which the hyperbola opens. It stretches through the vertices and crosses the center. For a vertical hyperbola like the one in our exercise, the transverse axis is vertical and thus coincides with the \(y\)-axis. Key aspects of the transverse axis include:

• Determining the length, which is \(2a\) where \(a\) is the distance from the center to a vertex.
• Providing understanding of the symmetry and orientation of the hyperbola.

This axis is important as it aids in plotting the hyperbola correctly and understanding its expansion direction.

Standard Form of Hyperbola

The standard form of a hyperbola's equation is a structure that provides comprehensive information about the shape and orientation of the curve. It is defined differently based on whether the hyperbola is vertical or horizontal. For a vertical hyperbola, like in our scenario, the standard form is:\[ \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \]This tells us:

• \(a^2\) corresponds to the distance to the vertices and thus influences the vertical stretch.
• \(b^2\) governs the hyperbola's horizontal configuration, affecting its width.

By inserting the known values into this format, one can precisely determine the specific equation that describes the hyperbolic shape fitting the given conditions. This ensures that the resulting equation is accurate and effectively represents the geometric properties of the hyperbola.