Problem 34 - Algebra and Trigonometry

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

The equation of the hyperbola is \(\frac{y^2}{36} - \frac{x^2}{20} = 1\).

Step 1: Identifying the Center and Vertices

The vertices are given as \((0, \pm 6)\). This indicates that the hyperbola is vertical, with the center at \((0, 0)\) and the vertices at \((0, 6)\) and \((0, -6)\). The distance from the center to each vertex is \(a = 6\).

Step 2: Write the Standard Equation of the Hyperbola

For a vertical hyperbola centered at \((0, 0)\), the standard form of the equation is \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). Since \(a = 6\), we have \(\frac{y^2}{36} - \frac{x^2}{b^2} = 1\).

Step 3: Using the Point to Find \(b^2\)

The hyperbola passes through the point \((-5, 9)\). Substitute \(x = -5\) and \(y = 9\) into the equation: \(\frac{9^2}{36} - \frac{(-5)^2}{b^2} = 1\).

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

Calculate \(\frac{9^2}{36} = \frac{81}{36} = 2.25\). Then the equation becomes \(2.25 - \frac{25}{b^2} = 1\). Solving for \(b^2\), we get \(2.25 - 1 = \frac{25}{b^2}\), so \(1.25 = \frac{25}{b^2}\). Thus, \(b^2 = \frac{25}{1.25} = 20\).

Step 5: Writing the Final Equation

Substitute \(b^2 = 20\) back into the standard equation: \(\frac{y^2}{36} - \frac{x^2}{20} = 1\). This is the equation of the hyperbola.

Key Concepts

Equation of Hyperbola

In algebra, hyperbolas are fascinating curves that emerge from the interaction of a plane and a double cone. Their distinctive shape is defined by a particular form of equation. Finding the equation of a hyperbola involves analyzing the given conditions—such as its vertices and a point through which it passes—to accurately represent the shape.
To find an equation for a hyperbola, we need certain components: the center, vertices, and sometimes other points on the curve. By evaluating these elements, you can construct a precise mathematical representation using its equation. This equation takes on different forms based on the hyperbola's orientation and position.

Standard Form

The standard form of the equation of a hyperbola provides a structured template from which calculations begin. For hyperbolas centered at the origin, the form depends on its orientation:

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

Here, \(a\) and \(b\) are distances from the center to the vertices and co-vertices respectively. This systematic framework helps simplify solving for unknown terms in the equation, aligning it correctly with its geometric properties.

Vertices

Vertices are key points on a hyperbola denoting its maximum excursion from the center along the major axis. For example, with given vertices at \((0, \pm 6)\), it signifies the hyperbola's major axis is vertical, with distance \(a\) being the length from center to vertex.
The vertices effectively establish the hyperbola's orientation and center location. By identifying the vertices, one can determine whether the hyperbola opens upwards and downwards (vertical case), or sideways (horizontal case), guiding the initial setup of the equation.

Substitute Values

Substituting values into the hyperbola's equation is a crucial step to solve for unknowns. This involves replacing \(x\) and \(y\) in the standard equation with specific coordinates of a point that the hyperbola passes through.In the exercise, using point \((-5, 9)\) in the equation \(\frac{y^2}{36} - \frac{x^2}{b^2} = 1\) allows us to derive the missing term \(b^2\).
Solve by calculating stepwise:

• Substitute the known coordinates.
• Perform algebraic operations to isolate the unknown \(b^2\).

This substitution is essential for finalizing the unique equation of a hyperbola conforming to all given conditions.

Geometry

Hyperbolas are more than algebraic equations—they are geometric figures with symmetries and well-defined properties. Understanding their geometry can aid in comprehending their mathematical representations.
Key geometric features include:

• Two parts or branches.
• Asymptotes that guide their shape.
• Symmetrical axes along vertices.

A hyperbola's geometry is interlinked with its equation, reflecting the spatial relationships between its components such as center, vertices, and asymptotes, offering a visual and conceptual understanding of these intriguing curves.