Problem 57 - 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. $$\frac{(x-6)^{2}}{10}-\frac{y^{2}}{40}=1$$

Step-by-Step Solution

Answer

Answer: The eccentricity of the given hyperbola is \(\frac{5}{\sqrt{5}}\).

Step 1: Identifying the values of 'a' and 'b'

From the given hyperbola equation, we can see that \(a^{2} = 10\) and \(b^{2} = 40\). To find the values of \(a\) and \(b\), we need to take the square root of both numbers. Thus, $$ a = \sqrt{10} \quad \text{and} \quad b = \sqrt{40} $$

Step 2: Using the Eccentricity Formula

Now that we have the values of \(a\) and \(b\), we can calculate the eccentricity of the hyperbola using the given formula: $$ \text{Eccentricity} =\frac{\sqrt{a^{2}+b^{2}}}{a} $$

Step 3: Calculating the Eccentricity

Substitute the values of \(a\) and \(b\) obtained in Step 1 into the formula from Step 2: $$ \text{Eccentricity} =\frac{\sqrt{(\sqrt{10})^{2}+(\sqrt{40})^{2}}}{\sqrt{10}} $$ Simplify the expression: $$ \text{Eccentricity} =\frac{\sqrt{10+40}}{\sqrt{10}} $$ Further simplification: $$ \text{Eccentricity} =\frac{\sqrt{50}}{\sqrt{10}} $$

Step 4: Simplify the eccentricity

We can simplify the eccentricity expression by factoring out common factors: $$ \text{Eccentricity} =\frac{\sqrt{2\cdot 25}}{\sqrt{2\cdot 5}} $$ Now we can cancel out the common factor of \(\sqrt{2}\) in the numerator and denominator: $$ \text{Eccentricity} =\frac{\sqrt{25}}{\sqrt{5}} $$ Finally, simplify the square root: $$ \text{Eccentricity} =\frac{5}{\sqrt{5}} $$ Therefore, the eccentricity of the given hyperbola is \(\frac{5}{\sqrt{5}}\).

Key Concepts

Eccentricity: Understanding the Measure of How 'Un-Circular' a Hyperbola Is

The term eccentricity is a crucial component in explaining the shape of various conic sections, including hyperbolas. Eccentricity, commonly denoted by the letter 'e,' is essentially a measure of how much a conic section deviates from being circular.
In simpler terms, it describes how stretched out the shape is.

• For a circle, the eccentricity is zero, indicating it's perfectly round.
• A parabola has an eccentricity of exactly one.
• Ellipses have an eccentricity less than one.
• Hyperbolas, like the one in our problem, always have an eccentricity greater than one.

When it comes to hyperbolas, the eccentricity formula is defined as \( e = \frac{\sqrt{a^2 + b^2}}{a} \), where \(a\) and \(b\) are distances related to the geometry of the hyperbola. This value gives us a precise word on how elongated the hyperbola is. The larger the eccentricity, the more "stretched" the hyperbola appears. Understanding this concept is vital for visualizing conic sections in a plane. By calculating it, you can quickly estimate the nature of the curve without plotting each point.

Conic Sections: The Diverse Family of Curves

Conic sections are shapes formed by intersecting a plane with a double-napped cone. Imagine slicing through an ice cream cone. Depending on the angle of your cut, you get different shapes.
These are collectively known as conic sections: circles, ellipses, parabolas, and hyperbolas.

• Circles and ellipses: Formed when the plane cuts through the cone at an angle less than the cone's side angle.
• Parabolas: Occur when the plane is parallel to the cone's side.
• Hyperbolas: Arise when the plane intersects both nappes of the cone, resulting in two separate curves.

Each of these shapes has unique properties and equations that set them apart. Hyperbolas, in particular, have two disconnected curves known as branches. Their standard form is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \) or \( \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \).
These shapes play a crucial role in various fields, from physics to engineering, guiding us in analyzing motion, forces, and optics. Recognizing hyperbolas as part of this larger family helps understand how they relate to other geometric figures.

Algebraic Manipulation: Simplifying with Skill

Algebraic manipulation involves rearranging and simplifying expressions and equations to make them easier to work with. It is a fundamental skill in mathematics that allows for the solving and understanding of more complex problems.
In the context of hyperbolas, this often involves cleaning up expressions to reveal underlying properties like eccentricity, making them more interpretable.

• One common technique is factoring, where expressions are rewritten to reveal common elements. This allows terms to be canceled out in fractions.
• Another is reducing square roots to simpler forms, as seen in our example: \( \frac{\sqrt{50}}{\sqrt{10}} \) simplifies to \( \frac{\sqrt{25}}{\sqrt{5}} \), reducing further to \( \frac{5}{\sqrt{5}} \) after cancellation.

Proper algebraic manipulation is crucial for efficiently solving math problems and can simplify complex issues, making them manageable. By practicing these skills, you become more adept at recognizing patterns and easier solutions in algebraic expressions.