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Chapter 11: Problem 48

Find an equation for the hyperbola described. Graph the equation. Vertices at (1,-3) and (1,1)\(;\) asymptote the line \(y+1=\frac{3}{2}(x-1)\)

Short Answer

Expert verified
\[ \frac{(y+1)^2}{4} - \frac{9(x-1)^2}{16} = 1 \]

Step by step solution

01

Identify the vertices

The vertices of the hyperbola are given at (1, -3) and (1, 1). The midpoint of these vertices gives the center of the hyperbola. Calculate the midpoint: \[ \text{Center} = \frac{(1+1)}{2}, \frac{(-3+1)}{2} = \frac{2}{2}, \frac{-2}{2} = (1, -1) \]
02

Determine the distance between vertices

The distance between (1, -3) and (1, 1) is 4, giving the value of 2a (the distance between the vertices is 2a). Therefore, \[ a = 2 \]
03

Identify orientation and equation form

Since the vertices share the same x-coordinate, the hyperbola opens vertically. The form of the hyperbola equation is: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \] Substituting the center (h, k) = (1, -1): \[ \frac{(y+1)^2}{a^2} - \frac{(x-1)^2}{b^2} = 1 \]
04

Determine the relationship of asymptote

The given asymptote equation is \( y + 1 = \frac{3}{2}(x - 1) \). The slope of the asymptote \( \frac{3}{2} \) is equal to \( \frac{a}{b} \). Therefore, \[ \frac{a}{b} = \frac{3}{2} \] Knowing that \( a = 2 \), solve for \( b \): \[ b = \frac{2a}{3} = \frac{2 \times 2}{3} = \frac{4}{3} \]
05

Write the equation of the hyperbola

Substitute \( a^2 = 4 \) and \( b^2 = (\frac{4}{3})^2 = \frac{16}{9} \) into the hyperbola equation: \[ \frac{(y+1)^2}{4} - \frac{(x-1)^2}{\frac{16}{9}} = 1 \] Simplify the equation: \[ \frac{(y+1)^2}{4} - \frac{9(x-1)^2}{16} = 1 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

hyperbola vertices
The vertices of a hyperbola are essential points that help define its geometry. In a hyperbola, the vertices are the points where the hyperbola intersects its principal axis.
For this problem, the vertices are given as (1, -3) and (1, 1). Since the x-coordinates are the same, the hyperbola opens vertically.
This tells us that the equation for our hyperbola will place emphasis on the y-term.
asymptote analysis
The asymptotes of a hyperbola are straight lines that the hyperbola approaches as it extends towards infinity. They provide crucial information about the shape and orientation of the hyperbola.
For this exercise, the given asymptote equation is: \[y + 1 = \frac{3}{2}(x - 1)\]
This equation is in the slope-intercept form 'y = mx + b', where the slope (m) is \(\frac{3}{2}\). This slope is critically tied to the 'a' and 'b' values in our hyperbola's equation through the relationship \( \frac{a}{b} = \frac{3}{2} \).
hyperbola center calculation
Finding the center of a hyperbola involves identifying the midpoint between its vertices. This midpoint is essential as it helps in forming the hyperbola's equation and understanding its geometry.
Given vertices: (1,-3) and (1, 1), calculate the midpoint: \( \text{Center} = ( \frac{(1+1)}{2}, \frac{(-3+1)}{2}) = (1, -1) \).
Thus, the center of the hyperbola is at the point (1, -1).
hyperbola equation form
The standard form of a hyperbola that opens vertically is: \[ \frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1 \]
Where (h,k) is the center of the hyperbola. From our previous steps, we have determined the center (h, k) to be (1, -1). This changes our general form to: \[ \frac{(y+1)^2}{a^2} - \frac{(x-1)^2}{b^2} = 1 \].
We further know that a=2 and that \( \frac{a}{b} = \frac{3}{2} \).
By substituting these values and performing the relevant calculations, we conclude with the hyperbola's equation in simplified form.
coordinate geometry
Understanding coordinate geometry is critical for analyzing and graphing hyperbolas.
  • The vertices give points of intersection with the principal axis.
  • The midpoints help find the center of the hyperbola.
  • Asymptotes provide information regarding orientation.
Combining these pieces, our hyperbola equation becomes: \[\frac{(y+1)^2}{4} - \frac{9(x-1)^2}{16} = 1\].
This equation reflects all characteristics, which can then be graphed for a visual representation.

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Most popular questions from this chapter

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Hyperbolic Mirrors Hyperbolas have interesting reflective properties that make them useful for lenses and mirrors. For example, if a ray of light strikes a convex hyperbolic mirror on a line that would (theoretically) pass through its rear focus, it is reflected through the front focus. This property, and that of the parabola, were used to develop the Cassegrain telescope in \(1672 .\) The focus of the parabolic mirror and the rear focus of the hyperbolic mirror are the same point. The rays are collected by the parabolic mirror, then reflected toward the (common) focus, and thus are reflected by the hyperbolic mirror through the opening to its front focus, where the eyepiece is located. If the equation of the hyperbola is \(\frac{y^{2}}{9}-\frac{x^{2}}{16}=1\) and the focal length (distance from the vertex to the focus) of the parabola is \(6,\) find the equation of the parabola.
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