Math Problem Statement

Determine the equation of the hyperbola with vertices (3, -1) and (3, -11) and asymptotes y = x - 9 and y = -x - 3.

Solution

To determine the equation of the hyperbola, we need to utilize the given information:

  • Vertices: (3,1)(3, -1) and (3,11)(3, -11)
  • Asymptotes: y=x9y = x - 9 and y=x3y = -x - 3

Step 1: Identify the center

The vertices have the same xx-coordinate, which indicates a vertical hyperbola (opening upwards and downwards). The center of the hyperbola is the midpoint of the two vertices.

  • Vertices: (3,1)(3, -1) and (3,11)(3, -11)
  • Midpoint (center): (3+32,1+(11)2)=(3,6)\left( \frac{3+3}{2}, \frac{-1+(-11)}{2} \right) = (3, -6)

Thus, the center of the hyperbola is (3,6)(3, -6).

Step 2: Equation of the asymptotes

For a vertical hyperbola, the asymptotes are given by the general form: yk=±ab(xh)y - k = \pm \frac{a}{b} (x - h) where (h,k)(h, k) is the center, aa is the distance from the center to the vertices (along the yy-axis), and bb is the distance from the center to the points where the asymptotes intersect the axes (along the xx-axis).

We are given two asymptotes: y=x9andy=x3y = x - 9 \quad \text{and} \quad y = -x - 3 These asymptotes have slopes of 11 and 1-1, respectively, which tells us that ab=1\frac{a}{b} = 1, so a=ba = b.

Step 3: Calculate aa (and thus bb)

The distance between the center (3,6)(3, -6) and each vertex is the value of aa. Since the vertices are located at (3,1)(3, -1) and (3,11)(3, -11), the distance between the center and the vertices is: a=1(6)=5a = \left| -1 - (-6) \right| = 5 Thus, a=5a = 5. Since a=ba = b, we also have b=5b = 5.

Step 4: Write the equation of the hyperbola

For a vertical hyperbola, the standard form is: (yk)2a2(xh)2b2=1\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1 Substituting (h,k)=(3,6)(h, k) = (3, -6), a=5a = 5, and b=5b = 5, the equation becomes: (y+6)225(x3)225=1\frac{(y + 6)^2}{25} - \frac{(x - 3)^2}{25} = 1 or simplifying: (y+6)225(x3)225=1\frac{(y + 6)^2}{25} - \frac{(x - 3)^2}{25} = 1

This is the equation of the hyperbola.

Would you like more details or clarification? Here are some related questions you could explore:

  1. How do you derive the equation of a hyperbola from different types of given data?
  2. Why do the asymptotes for hyperbolas have the form they do?
  3. What is the significance of the values of aa and bb in a hyperbola?
  4. How do horizontal and vertical hyperbolas differ in their general form?
  5. How would the equation change if the hyperbola had a different center?

Tip: Always ensure to correctly identify the center and axes orientation (horizontal or vertical) when working with conic sections like hyperbolas.

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Math Problem Analysis

Mathematical Concepts

Hyperbolas
Conic Sections
Analytical Geometry

Formulas

Midpoint formula: Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
Equation of a hyperbola: (y - k)^2/a^2 - (x - h)^2/b^2 = 1
Asymptotes of a hyperbola: y - k = ± (a/b)(x - h)

Theorems

Properties of Hyperbolas
Equations of Asymptotes

Suitable Grade Level

Grades 10-12