Algebra 2 Hyperbola Equation Problems with Detailed Solutions

📚 Understanding Hyperbolas: A Comprehensive Guide

A hyperbola is a type of conic section formed by the intersection of a plane with a double cone. It consists of two separate curves (branches) that are mirror images of each other. The equation of a hyperbola describes the relationship between the x and y coordinates of points on these curves.

• 🔍 Definition: A hyperbola is the locus of a point which moves so that the difference of its distances from two fixed points (foci) is constant.
• 📜 History: Hyperbolas were first studied by Menaechmus in his investigation of the Delian problem. Euclid and Aristaeus wrote about hyperbolas, but Apollonius gave the curve its name.

📐 Key Principles of Hyperbola Equations

The standard form of a hyperbola equation depends on whether it opens horizontally or vertically and is centered at the origin (0,0) or at a point (h,k). Understanding these forms is crucial for solving problems.

• 📈 Horizontal Hyperbola (Centered at Origin): The equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. The vertices are at $(\pm a, 0)$.
• 📉 Vertical Hyperbola (Centered at Origin): The equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$. The vertices are at $(0, \pm a)$.
• 📍 Horizontal Hyperbola (Centered at (h,k)): The equation is $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$.
• 📌 Vertical Hyperbola (Centered at (h,k)): The equation is $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$.
• 🔗 Relationship between a, b, and c: $c^2 = a^2 + b^2$, where 'c' is the distance from the center to each focus.
• 🧭 Asymptotes: These are lines that the hyperbola approaches as it extends to infinity. For a hyperbola centered at the origin, the asymptotes are $y = \pm \frac{b}{a}x$ (horizontal) and $y = \pm \frac{a}{b}x$ (vertical). For hyperbolas centered at (h,k), adjust accordingly using point-slope form.

💡 Example Problems with Detailed Solutions

Let's work through some common algebra 2 hyperbola problems.

1. Problem 1: Find the equation of a hyperbola centered at the origin with a vertex at (5, 0) and a focus at (13, 0).
   Solution: Since the vertex is at (5, 0), we know $a = 5$. The focus is at (13, 0), so $c = 13$. Using $c^2 = a^2 + b^2$, we have $13^2 = 5^2 + b^2$, which gives $b^2 = 169 - 25 = 144$, so $b = 12$. The equation is $\frac{x^2}{25} - \frac{y^2}{144} = 1$.
2. Problem 2: Find the equation of a hyperbola with vertices at (0, 4) and (0, -4) and foci at (0, 6) and (0, -6).
   Solution: This is a vertical hyperbola. Since the vertices are at (0, $\pm$4), $a = 4$. The foci are at (0, $\pm$6), so $c = 6$. Using $c^2 = a^2 + b^2$, we have $6^2 = 4^2 + b^2$, which gives $b^2 = 36 - 16 = 20$. The equation is $\frac{y^2}{16} - \frac{x^2}{20} = 1$.
3. Problem 3: Write the equation of the hyperbola $\frac{(x-2)^2}{9} - \frac{(y+1)^2}{16} = 1$ in standard form and identify the center, vertices, and foci.
   Solution: The equation is already in standard form. The center is (2, -1). Here, $a^2 = 9$ so $a = 3$, and $b^2 = 16$ so $b = 4$. $c^2 = a^2 + b^2 = 9 + 16 = 25$, so $c = 5$. Since it's a horizontal hyperbola, the vertices are (2$\pm$3, -1) which are (5, -1) and (-1, -1). The foci are (2$\pm$5, -1) which are (7, -1) and (-3, -1).
4. Problem 4: Determine the equation of the hyperbola with center at (-3, 2), a focus at (2, 2) and vertex at (-1, 2).
   Solution: Since the center is (-3, 2) and the vertex is (-1, 2) then a = |-1 - (-3)| = 2. Since the focus is at (2, 2), c = |2 - (-3)| = 5. Therefore $b^2 = c^2 - a^2 = 25 - 4 = 21$. Since the foci and vertices share the same y-coordinate, this is a horizontal hyperbola with equation: $\frac{(x+3)^2}{4} - \frac{(y-2)^2}{21} = 1$.
5. Problem 5: Find the equation of the asymptotes of the hyperbola $\frac{y^2}{4} - \frac{x^2}{9} = 1$.
   Solution: This is a vertical hyperbola centered at the origin. Here, $a^2 = 4$ so $a = 2$, and $b^2 = 9$ so $b = 3$. The asymptotes are $y = \pm \frac{a}{b}x = \pm \frac{2}{3}x$.
6. Problem 6: Given the equation $9x^2 - 16y^2 - 36x - 96y - 252 = 0$, rewrite the equation in standard form and identify the center.
   Solution: Complete the square for both x and y terms: $9(x^2 - 4x) - 16(y^2 + 6y) = 252$. $9(x^2 - 4x + 4) - 16(y^2 + 6y + 9) = 252 + 9(4) - 16(9)$. $9(x - 2)^2 - 16(y + 3)^2 = 252 + 36 - 144 = 144$. $\frac{(x - 2)^2}{16} - \frac{(y + 3)^2}{9} = 1$. The center is (2, -3).
7. Problem 7: Identify the key features (center, vertices, foci, asymptotes) of the hyperbola defined by $\frac{(x+1)^2}{25} - \frac{(y-2)^2}{9} = 1$.
   Solution: This is a horizontal hyperbola. The center is (-1, 2). $a^2 = 25$ so $a = 5$, and $b^2 = 9$ so $b = 3$. $c^2 = a^2 + b^2 = 25 + 9 = 34$, so $c = \sqrt{34}$. The vertices are (-1$\pm$5, 2) which are (4, 2) and (-6, 2). The foci are (-1$\pm\sqrt{34}, 2)$. The asymptotes are $y - 2 = \pm \frac{3}{5}(x + 1)$.

🌍 Real-world Applications

Hyperbolas appear in various real-world applications, including:

• 📡 Satellite Dishes: The cross-section of a satellite dish is often a parabola or a hyperbola.
• 🛰️ Navigation Systems: LORAN (Long Range Navigation) uses hyperbolas to determine the location of a ship or aircraft.
• 🔭 Telescopes: Some telescope designs incorporate hyperbolic mirrors.

🧠 Conclusion

Understanding the equation of a hyperbola and its key components is essential for success in Algebra 2. By mastering the standard forms and practicing problem-solving techniques, you can confidently tackle any hyperbola-related challenge. Keep practicing, and you'll become a hyperbola expert! 🚀