Problem 60 - Contemporary Precalculus

Question

If \(a>0\) and \(b>0,\) then the eccentricity of the hyperbola $$\frac{(x-h)^{2}}{a^{2}}-\frac{(y-k)^{2}}{b^{2}}=1 \quad \text { or } \quad \frac{(y-k)^{2}}{a^{2}}-\frac{(x-h)^{2}}{b^{2}}=1$$ is the number \(\frac{\sqrt{a^{2}+b^{2}}}{a} .\) Find the eccentricity of the hyperbola whose equation is given. $$16 x^{2}-9 y^{2}-32 x+36 y+124=0$$

Step-by-Step Solution

Answer

Answer: The eccentricity of the hyperbola is \(\frac{\sqrt{73}}{4}\).

Step 1: Rewrite the equation in standard form

First, we group the terms containing \(x\) and \(y\): $$16(x^2 - 2x) - 9(y^2 - 4y) = -124$$ Now, complete the square for both the \(x\) and \(y\) terms: $$(x-1)^2 = x^2-2x+1$$ $$(y+2)^2 = y^2+4y+4$$ Hence, we can rewrite the given equation as: $$16[(x-1)^2 - 1] - 9[(y+2)^2-4]=-124$$ Simplify the equation to obtain the standard form: $$\frac{(x-1)^2}{(1/16)} - \frac{(y+2)^2}{(4/9)} = 1$$ Now, we can identify \(a^2\) and \(b^2\): $$a^2 = \frac{1}{16}$$ $$b^2 = \frac{4}{9}$$

Step 2: Find the eccentricity of the hyperbola

Using the formula for the eccentricity of the hyperbola given in the exercise: $$\text{eccentricity} = \frac{\sqrt{a^2+b^2}}{a}$$ Substitute the values of \(a^2\) and \(b^2\) into the formula: $$\text{eccentricity} = \frac{\sqrt{\frac{1}{16}+\frac{4}{9}}}{\frac{1}{4}}$$ Simplify the expression: $$\text{eccentricity} = \frac{\sqrt{\frac{73}{144}}}{\frac{1}{4}}=\frac{1}{4}\cdot\frac{\sqrt{73}}{\sqrt{1/16}}$$ $$\text{eccentricity} = \frac{\sqrt{73}}{4}$$ Hence, the eccentricity of the hyperbola with the given equation is \(\frac{\sqrt{73}}{4}\).

Key Concepts

Standard Form of Hyperbola

The equation of a hyperbola can be expressed in what we call the "standard form." This means rewriting the equation to clearly identify its structure. Hyperbolas have two primary forms, based on their orientation:

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

For either form, you need to find values for \(a^2\) and \(b^2\). These determine the hyperbola's shape and size. The center of the hyperbola is \((h, k)\). In solving these equations, properly transforming the initial equation into the standard form is key. It sets up all subsequent mathematical operations, like finding the eccentricity. Whether it starts from a general form or requires rearranging terms, the standard form provides clarity for further computations.

Completing the Square

Completing the square is a crucial mathematical technique used for rewriting quadratic expressions in a more workable form. Often used in conic sections, it allows us to identify and isolate terms that reveal essential features such as the center of a hyperbola. Here’s how to complete the square:

• Identify the quadratic terms from the equation, like \(x^2 + 2bx\) or \(y^2 + 2cy\).
• Separate the quadratic and linear part: \((x + b)^2 - b^2\).
• Add and subtract the same square number to balance the equation.

In the initial exercise, completing the square was applied to both \(x\) and \(y\) terms separately. This process simplifies the quadratic equation, making it easier to match the standard form previously discussed. Completing the square transforms such expressions into neat squares, aligning them perfectly with the required form.

Polynomials Simplification

When dealing with hyperbola equations, simplifying polynomials is a necessary step. This involves manipulating the equation, either by combining like terms or by simplifying fractions and radical expressions. In the case of our exercise, once the squares were completed, rearranging terms effectively simplified the equation:

• Consolidate and organize terms to reveal factors of the polynomial.
• Simplify these factors as fractions, radicals, or other algebraic expressions to solve for key values like \(a^2\) and \(b^2\).
• Recognize and eliminate unnecessary parts of the equation.

Polynomials simplification helps in identifying relevant components for the hyperbola's standard form. This is essential in preparing the equation for solving or further manipulation, such as determining its eccentricity.

Learn Precalculus

Precalculus is an introductory course that covers fundamental concepts in preparation for calculus. It includes topics such as algebraic expressions, functions, analytical geometry, and trigonometry. Here, we delve into the concepts of conic sections, including hyperbolas, which are key components of precalculus courses.

• Understanding how to transform equations into different forms, like the standard form, equips students with problem-solving skills.
• Techniques like completing the square can be applied broadly across various mathematical problems and disciplines.
• Grasping these concepts enhances one's ability to manipulate complex equations and promote a deeper understanding of functions and their characteristics.

Studying precalculus provides the necessary toolkit to tackle calculus with confidence. It's about building solid mathematical foundations and refining analytical skills.