Solved Examples: Writing Hyperbola Equations Given Vertices and Asymptote Lines

📚 Quick Study Guide

• 🧭 Definition: A hyperbola is a conic section formed by the intersection of a double cone with a plane.
• 🍎 Standard Equation (Horizontal Transverse Axis): $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$, where $(h, k)$ is the center.
• 🍇 Standard Equation (Vertical Transverse Axis): $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$, where $(h, k)$ is the center.
• 🥬 Vertices: Located at $(h \pm a, k)$ for a horizontal transverse axis, and $(h, k \pm a)$ for a vertical transverse axis.
• 🍋 Asymptotes (Horizontal Transverse Axis): $y - k = \pm \frac{b}{a}(x - h)$
• 🥝 Asymptotes (Vertical Transverse Axis): $y - k = \pm \frac{a}{b}(x - h)$
• 💡 Key Relationship: $c^2 = a^2 + b^2$, where $c$ is the distance from the center to each focus.
• 📐 Finding the Equation: Use the given vertices to find the center $(h, k)$ and the value of $a$. Then, use the asymptotes' slopes to find the value of $b$. Finally, plug $a$, $b$, $h$, and $k$ into the appropriate standard equation.

Practice Quiz

1. The vertices of a hyperbola are at $(-1, 2)$ and $(5, 2)$, and its asymptotes have slopes of $\pm \frac{2}{3}$. What is the equation of the hyperbola?

   1. $\frac{(x-2)^2}{9} - \frac{(y-2)^2}{4} = 1$
   2. $\frac{(x-2)^2}{4} - \frac{(y-2)^2}{9} = 1$
   3. $\frac{(y-2)^2}{9} - \frac{(x-2)^2}{4} = 1$
   4. $\frac{(y-2)^2}{4} - \frac{(x-2)^2}{9} = 1$

2. A hyperbola has vertices at $(3, -1)$ and $(3, 5)$. Its asymptotes are $y + 1 = \pm \frac{3}{2}(x - 3)$. What is the equation of the hyperbola?

   1. $\frac{(y-2)^2}{9} - \frac{(x-3)^2}{4} = 1$
   2. $\frac{(y-2)^2}{4} - \frac{(x-3)^2}{9} = 1$
   3. $\frac{(x-3)^2}{9} - \frac{(y-2)^2}{4} = 1$
   4. $\frac{(x-3)^2}{4} - \frac{(y-2)^2}{9} = 1$

3. The vertices of a hyperbola are at $(-2, -3)$ and $(-2, 7)$, and the asymptotes have a slope of $\pm 2$. What is the equation of this hyperbola?

   1. $\frac{(y-2)^2}{25} - \frac{(x+2)^2}{100} = 1$
   2. $\frac{(y-2)^2}{25} - \frac{(x+2)^2}{\frac{25}{4}} = 1$
   3. $\frac{(x+2)^2}{25} - \frac{(y-2)^2}{\frac{25}{4}} = 1$
   4. $\frac{(x+2)^2}{25} - \frac{(y-2)^2}{100} = 1$

4. A hyperbola's vertices are at $(1, 1)$ and $(7, 1)$, and its asymptotes are $y - 1 = \pm \frac{1}{3}(x - 4)$. What is the equation of the hyperbola?

   1. $\frac{(x-4)^2}{9} - \frac{(y-1)^2}{1} = 1$
   2. $\frac{(x-4)^2}{1} - \frac{(y-1)^2}{9} = 1$
   3. $\frac{(y-1)^2}{9} - \frac{(x-4)^2}{1} = 1$
   4. $\frac{(y-1)^2}{1} - \frac{(x-4)^2}{9} = 1$

5. What is the center of a hyperbola with vertices at $(-5, 0)$ and $(5, 0)$?

   1. $(0, 0)$
   2. $(5, 0)$
   3. $(-5, 0)$
   4. $(0, 5)$

6. If a hyperbola has vertices at $(0, -3)$ and $(0, 3)$ and asymptotes $y = \pm x$, what is the value of $b$ in its equation?

   1. 3
   2. 6
   3. 9
   4. 1

7. A hyperbola has vertices at $(-2, 1)$ and $(6, 1)$. One of its asymptotes is $y = \frac{-1}{2}x + 2$. Find the equation of the hyperbola.

   1. $\frac{(x-2)^2}{16} - \frac{(y-1)^2}{4} = 1$
   2. $\frac{(x-2)^2}{16} - \frac{(y-1)^2}{64} = 1$
   3. $\frac{(x+2)^2}{16} - \frac{(y+1)^2}{4} = 1$
   4. $\frac{(x-2)^2}{4} - \frac{(y-1)^2}{16} = 1$