SOLUTION: Find the center, vertices, foci, and asymptotes of the hyperbola 4x^2-y^2-8x-2y-1=0. Solve and graph

Question 1194363: Find the center, vertices, foci, and asymptotes of the hyperbola 4x^2-y^2-8x-2y-1=0. Solve and graph
Answer by asinus(45) (Show Source): You can put this solution on YOUR website!
To analyze and graph the hyperbola given by \( 4x^2 - y^2 - 8x - 12y - 1 = 0 \), we must rewrite it in the standard form. Here are the steps to solve this problem:
---
### **Solution By Steps**
---
#### **Step 1: Rewrite the equation in standard form**
1. Group \(x\)-terms and \(y\)-terms:
\[
4x^2 - 8x - y^2 - 12y = 1
\]
2. Factor out coefficients of \(x^2\) and \(y^2\) if needed, and complete the square:
\[
4(x^2 - 2x) - (y^2 + 12y) = 1
\]
- For \(4(x^2 - 2x)\):
The term to complete the square is \((\frac{-2}{2})^2 = 1\).
- For \(-(y^2 + 12y)\):
The term to complete the square is \((\frac{12}{2})^2 = 36\).
3. Add these terms inside the parentheses and balance the equation:
\[
4(x^2 - 2x + 1) - (y^2 + 12y + 36) = 1 + 4(1) - 36
\]
Simplify:
\[
4(x - 1)^2 - (y + 6)^2 = -31
\]
#### **Step 2: Transform into standard form**
1. Divide through by \(-31\) to standardize the equation:
\[
\frac{(y + 6)^2}{31} - \frac{(x - 1)^2}{31/4} = 1
\]
2. Rewrite to make the signs and denominators clear:
\[
\frac{(y + 6)^2}{31} - \frac{(x - 1)^2}{\frac{31}{4}} = 1
\]
Thus, the hyperbola is **vertical** with the center at \((1, -6)\).
---
#### **Step 3: Determine the key characteristics**
1. **Center**: \((1, -6)\)
2. **Vertices**: These are located \(a = \sqrt{31}\) units above and below the center since it is a vertical hyperbola.
- Vertices: \((1, -6 + \sqrt{31})\) and \((1, -6 - \sqrt{31})\)
3. **Foci**: The distance to the foci is given by \(c = \sqrt{a^2 + b^2} = \sqrt{31 + \frac{31}{4}} = \sqrt{\frac{155}{4}} = \frac{\sqrt{155}}{2}\).
- Foci: \((1, -6 + \frac{\sqrt{155}}{2})\) and \((1, -6 - \frac{\sqrt{155}}{2})\)
4. **Asymptotes**: The equations for the asymptotes are:
\[
y - (-6) = \pm \frac{\sqrt{\frac{31}{4}}}{\sqrt{31}} (x - 1)
\]
Simplify the slopes:
\[
y + 6 = \pm \frac{1}{2}(x - 1)
\]
---
### **Final Answer**
- **Center**: \((1, -6)\)
- **Vertices**: \((1, -6 + \sqrt{31})\), \((1, -6 - \sqrt{31})\)
- **Foci**: \((1, -6 + \frac{\sqrt{155}}{2})\), \((1, -6 - \frac{\sqrt{155}}{2})\)
- **Asymptotes**: \(y + 6 = \frac{1}{2}(x - 1)\), \(y + 6 = -\frac{1}{2}(x - 1)\)
---
### **Key Concept**
A hyperbola is defined as the set of points where the absolute difference of distances to two fixed points (foci) is constant.
---
### **Key Concept Explanation**
1. **Standard Form of Hyperbola**:
- Vertical hyperbola: \(\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1\)
- Horizontal hyperbola: \(\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1\)
2. **Key Properties**:
- The **center** is \((h, k)\).
- The **vertices** are \(a\) units from the center along the major axis.
- The **foci** are \(c = \sqrt{a^2 + b^2}\) units from the center along the major axis.
- The **asymptotes** guide the hyperbola's approach to infinity and are determined by the slopes \(\pm \frac{a}{b}\).
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