Problem 61 - 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. $$4 x^{2}-5 y^{2}-16 x-50 y+71=0$$

Step-by-Step Solution

Answer

Answer: The eccentricity of the given hyperbola is \(\frac{\sqrt{29}}{5}\).

Step 1: Move constants to the right and rearrange the equation

In order to rewrite the given equation into the standard form, we should first move the constant 71 to the right side of the equation, so it becomes: $$4x^2 - 5y^2 - 16x - 50y = -71$$ Now, let's rearrange the equation by separating the terms with \(x\) and \(y\): $$4x^2 - 16x - 5y^2 - 50y= -71$$

Step 2: Complete the square for both x and y

To rewrite this equation in terms of a standard hyperbolic equation, we need to complete the squares for both the x and y terms. To do that, we will be factoring out the coefficients for the square terms and add the required constants in both sides of the equation: $$4(x^2 - 4x) - 5(y^2 + 10y) = -71$$ $$4[(x^2 - 4x + 4) - 4] - 5[(y^2 + 10y + 25) - 25] = -71$$ Now simplify the equation: $$4[(x-2)^2 - 4] - 5[(y+5)^2 - 25] = -71$$

Step 3: Convert into the standard hyperbolic equation form

Now we will divide the equation by -1 and move the constants to the other side: $$5[(y+5)^2 - 25] - 4[(x-2)^2 - 4] = 71$$ Next, rewrite the equation in terms of fractions: $$\frac{(y+5)^2}{25} - \frac{(x-2)^2}{4} = 1$$ Now our equation is in the standard hyperbolic form. Comparing it with the given standard equations, we can see it matches the second one, so we have: $$\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$$ By looking at the matching terms, we can find the values of \(a\) and \(b\) as: $$a^2 = 25 \Rightarrow a = 5$$ $$b^2 = 4 \Rightarrow b = 2$$

Step 4: Calculate the eccentricity

Now we have the values of \(a\) and \(b\). Using the given formula for the eccentricity, we can find the eccentricity of the hyperbola as follows: $$e = \frac{\sqrt{a^2+b^2}}{a}$$ Plugging in the values of \(a\) and \(b\): $$e = \frac{\sqrt{5^2 + 2^2}}{5}$$ $$e = \frac{\sqrt{25+4}}{5}$$ $$e = \frac{\sqrt{29}}{5}$$ So, the eccentricity of the given hyperbola is \(\frac{\sqrt{29}}{5}\).

Key Concepts

Eccentricity

The eccentricity of a hyperbola is an important dimension that helps to define the shape of the hyperbola. The eccentricity, denoted typically by the letter \(e\), is a measure of how much a conic section deviates from being circular.
For hyperbolas, the eccentricity is always greater than 1. The formula to find the eccentricity for the hyperbola given by the equation \[ \frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1 \] is given by:

• \( e = \frac{\sqrt{a^2 + b^2}}{a} \)

The greater the eccentricity, the more 'stretched out' the hyperbola looks. This is because the foci, which are points that the hyperbola revolves around, become further apart as eccentricity increases. In simpler terms, with higher eccentricity, the arms of the hyperbola spread further apart.
Using our given example, where \( a = 5 \) and \( b = 2 \), the eccentricity would be calculated as \( \frac{\sqrt{29}}{5} \), indicating that our hyperbola is reasonably 'stretched out', but still maintains its characteristic hyperbolic path.

Completing the square

Completing the square is a method used to transform a quadratic equation into a perfect square trinomial plus or minus a constant. This is extremely useful in conic sections because it helps us to derive the standard form of equations like that of a hyperbola or an ellipse. Here's how it's generally done:

• Given a quadratic expression \( ax^2 + bx + c \), focus on the terms with \(x\).
• Factor out the coefficient of \(x^2\) if it's not 1.
• Add and subtract \((\frac{b}{2a})^2\) inside the parentheses to form a perfect square trinomial.

For example, consider transforming \(4(x^2 - 4x)\) into a perfect square. First, find the square to complete: \((\frac{-4}{2})^2 = 4\). Adding and subtracting \(4\) inside the parentheses gives you:\[4((x^2 - 4x + 4) - 4) = 4((x-2)^2 - 4) \]
We also do this for the y-terms in our hyperbola, enabling us to manipulate the equation into the standard form. The method is not just a trick but a solid technique to systematically solve and convert quadratic equations in algebra into more recognizable forms.

Conic sections

Conic sections are curves obtained by cutting a cone with a plane. These include circles, ellipses, parabolas, and hyperbolas. They appear frequently in geometry, algebra, and real-world applications. The hyperbola is one such section defined as the set of points wherein the difference of the distances to two foci is constant.
In mathematical terms, a hyperbola looks like two mirrored arcs that open either horizontally or vertically. The general equation of a hyperbola is:

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

This form tells you the orientation and shape of the hyperbola. The center of the hyperbola is \((h, k)\), \(a\) represents the distance from the center to the vertices along the axis of symmetry, and \(b\) represents the distance along the other axis.
In any conic section, altering values like \(a\) and \(b\) changes the shape. Understanding conic sections is essential because hyperbolas, ellipses, and parabolas are fundamental shapes that describe planetary orbits, reflectivity properties, and more in math and science.