Problem 42 - Contemporary Precalculus

Question

Find the equation of the hyperbola that satisfies the given conditions. Center (-3,-5)\(;\) vertex (-3,0)\(;\) asymptote \(6 y=5 x-15\)

Step-by-Step Solution

Answer

Answer: The equation of the hyperbola is \((\frac{(x+3)^2}{25}) - (\frac{(y+5)^2}{61}) = 1\).

Step 1: Determine the values of a, h, and k

We know that the center of the hyperbola is at (-3,-5), so h = -3 and k = -5. We also know that one of the vertices is at (-3, 0), so to find the value of `a`, we need to find the distance between the center and this vertex, which represents half the distance between the vertices of the hyperbola along its transverse axis. The distance between the center and the vertex = \(|0-(-5)| = |5| = 5\) So, a = 5, h = -3, and k = -5.

Step 2: Determine the slope and equation of the given asymptote

We are given the equation of one asymptote: \(6 y = 5 x - 15\). First, we rewrite this as a standard linear equation: y = \(\frac{5}{6} x -\frac{15}{6}\) Now, we can see that the slope of the given asymptote is m1 = \(\frac{5}{6}\).

Step 3: Calculate the slope of the other asymptote

Since the asymptotes of a hyperbola are perpendicular to each other at the center, the product of their slopes is equal to -1. Therefore, we can determine the slope of the other asymptote using the formula m1 * m2 = -1: m2 = \(\frac{-1}{m1}\) = \(\frac{-1}{\frac{5}{6}}\) = \(\frac{-6}{5}\)

Step 4: Determine the equation of the second asymptote

As the equation of one asymptote passing through center is determined, we can now find the equation of the other asymptote in the form y = m2 * (x - h) + k, by substituting the values of h, k, and m2: y = \(\left(\frac{-6}{5}\right)\) * (x - (-3)) + (-5) y = \(\frac{-6}{5}\)(x + 3) - 5

Step 5: Calculate the value of b

The distance between the intersection points of the asymptotes and the hyperbola is the same as their horizontal and vertical distances between the center and the vertices. Therefore, we can find the intersection points on the hyperbola's conjugate axis for the given asymptote equations: y = \(\frac{5}{6}\)(x + 3) - 5, and y = \(\frac{-6}{5}\)(x + 3) - 5 The intersection points on the given line equations are found by equating the two line equations: \(\frac{5}{6}\)(x + 3) - 5 = \(\frac{-6}{5}\)(x + 3) - 5 We can now solve the equation to find the x-coordinate of the intersection point: \(x(\frac{5}{6} + \frac{-6}{5}) = 0\) Therefore, x = 3. Now, we find the y-coordinate of the intersection point by substituting x = 3 into one of the asymptote equations: y = \(\frac{5}{6}\)(3 + 3) - 5 y = \(\frac{5}{6}\) * 6 - 5 y = 5 - 5 y = 0 The intersection point of the lines is (3, 0). Now, we find the distance between the center and this intersection point, which represents the value of b: b = sqrt((3 - (-3))^2 + (0 - (-5))^2) = sqrt(6^2 + 5^2) = sqrt(61)

Step 6: Write the equation of the hyperbola

Now that we have the values of a, b, h, and k, we can write the equation of the hyperbola in its standard form: \((\frac{(x-h)^2}{a^2}) - (\frac{(y-k)^2}{b^2}) = 1\) By substituting the values we found earlier: \((\frac{(x-(-3))^2}{5^2}) - (\frac{(y-(-5))^2}{61}) = 1\) The equation of the hyperbola is: \((\frac{(x+3)^2}{25}) - (\frac{(y+5)^2}{61}) = 1\)

Key Concepts

Center of Hyperbola

The center of a hyperbola is a crucial point from which the entire figure is symmetric. It's like the heart of the hyperbola.
Understanding its role helps you grasp the overall layout of the shape.

• The center is denoted as \((h, k)\).
• For a given hyperbola, the center is always situated at the midpoint of the space between the two vertices.
• In our example, the center is at \((-3, -5)\), indicating this is where the hyperbola pivots symmetrically.

This midpoint property is instrumental for setting up the entire equationand for orienting other key elements like vertices and asymptotes.

Vertices

Vertices of a hyperbola mark the juncture of maximum curvature on the figure's branches. They also help determine the size and the shape of the hyperbola.
Here's why understanding them is important:

• The vertices are always symmetrically placed about the center.
• The value of \(a\), representing half the distance between the vertices along the transverse axis, is calculated using the distance formula with the center.
• In this case, the vertex provided is \((-3, 0)\), leading to \(a = 5\) since the distance between the center \((-3, -5)\) and the vertex \((-3, 0)\) is 5 units.

This configuration helps establish the hyperbola's width horizontallywhen you plot out the equation.

Asymptotes

Asymptotes are powerful guides to the directional growth of hyperbola branches. They do not intersect the hyperbola but show its trajectory.
They are essential for sketching and understanding the framework of hyperbolas.

• For a hyperbola, the lines of asymptotes intersect at the center.
• In our example, the given asymptote is \(6y = 5x - 15\), equivalent to \(y = \frac{5}{6}x - 2.5\), with a slope \(\frac{5}{6}\).
• The second asymptote, due to perpendicularity (product of slopes equal to -1), is thus \(y = \frac{-6}{5}(x + 3) - 5\).

The asymptotes sketch a symmetric compensating slope that defines the open arms of the hyperbola, providing a glimpse at their infinite extension.

Distance Formula

The distance formula is a valuable tool in determining various aspects, like the span of a hyperbola between its center and vertices. It is crucial for solving many geometric problems.

• The formula calculates the distance between two points: \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\).
• This formula helps measure \(a\) when you find the distance between the center and vertices, as seen with \(|0 - (-5)| = 5\) for our hyperbola.
• Also, it is used to find \(b\) through distance between center and asymptote intersection.

Using the distance formula efficiently allows accurate and precise calculation of these crucial hyperbolic propertiesthat form the basis of the hyperbola's equation.

Conjugate Axis

The conjugate axis is perpendicular to the transverse axis and helps define the shape and proportion of a hyperbola. Knowing its role clarifies the positioning and structure of a hyperbola.

• The conjugate axis extends through the center of the hyperbola at right angles to the transverse axis.
• It correlates to the \(b\) parameter of the hyperbola, shaping its breadth.
• In our example, \(b\) is calculated as \(\sqrt{61}\) through intersection points along the asymptotes, reflecting the vertical orientation measure.

This axis, while not hosting the vertices, ensures the structural volumeand proportionality of the hyperbola's shape.