1. x2 + 100y² + 2x - 1000y + 2401 = 0
2. 9x² + 25 y² - 36x + 50y - 164 = 0

convert the following general form to standard form. then, identify the center, vertices, and co-vertices  the ellipse

To convert the given equations from their general form to the standard form of an ellipse, we need to complete the square for both the \(x\) and \(y\) terms.

### 1. Equation: \(x^2 + 100y^2 + 2x - 1000y + 2401 = 0\)

**Step 1: Group the \(x\) and \(y\) terms together.**
\[
x^2 + 2x + 100y^2 - 1000y + 2401 = 0
\]

**Step 2: Move the constant term to the other side.**
\[
x^2 + 2x + 100y^2 - 1000y = -2401
\]

**Step 3: Complete the square for the \(x\) and \(y\) terms.**

For \(x\):
\[
x^2 + 2x = (x + 1)^2 - 1
\]

For \(y\):
\[
100(y^2 - 10y) = 100\left[(y - 5)^2 - 25\right] = 100(y - 5)^2 - 2500
\]

**Step 4: Substitute these into the equation and simplify.**
\[
(x + 1)^2 - 1 + 100(y - 5)^2 - 2500 = -2401
\]
\[
(x + 1)^2 + 100(y - 5)^2 - 2501 = -2401
\]
\[
(x + 1)^2 + 100(y - 5)^2 = 100
\]

**Step 5: Divide the entire equation by 100 to get the standard form.**
\[
\frac{(x + 1)^2}{100} + \frac{(y - 5)^2}{1} = 1
\]

**Identifying the center, vertices, and co-vertices:**

- **Center:** \((-1, 5)\)
- **Vertices:** For the \(x\)-term (major axis), \( \pm 10 \) from the center:
  - \((-11, 5)\) and \((9, 5)\)
- **Co-vertices:** For the \(y\)-term (minor axis), \( \pm 1 \) from the center:
  - \((-1, 4)\) and \((-1, 6)\)

### 2. Equation: \(9x^2 + 25y^2 - 36x + 50y - 164 = 0\)

**Step 1: Group the \(x\) and \(y\) terms together.**
\[
9x^2 - 36x + 25y^2 + 50y = 164
\]

**Step 2: Factor out the coefficients of \(x^2\) and \(y^2\) terms.**
\[
9(x^2 - 4x) + 25(y^2 + 2y) = 164
\]

**Step 3: Complete the square for the \(x\) and \(y\) terms.**

For \(x\):
\[
x^2 - 4x = (x - 2)^2 - 4
\]

For \(y\):
\[
y^2 + 2y = (y + 1)^2 - 1
\]

**Step 4: Substitute these into the equation and simplify.**
\[
9[(x - 2)^2 - 4] + 25[(y + 1)^2 - 1] = 164
\]
\[
9(x - 2)^2 - 36 + 25(y + 1)^2 - 25 = 164
\]
\[
9(x - 2)^2 + 25(y + 1)^2 = 225
\]

**Step 5: Divide the entire equation by 225 to get the standard form.**
\[
\frac{(x - 2)^2}{25} + \frac{(y + 1)^2}{9} = 1
\]

**Identifying the center, vertices, and co-vertices:**

- **Center:** \((2, -1)\)
- **Vertices:** For the \(x\)-term (major axis), \( \pm 5 \) from the center:
  - \((-3, -1)\) and \((7, -1)\)
- **Co-vertices:** For the \(y\)-term (minor axis), \( \pm 3 \) from the center:
  - \((2, -4)\) and \((2, 2)\)

---

Would you like more details on any part of the process, or do you have any questions?

### Follow-up Questions:
1. How do you complete the square for different types of conic sections?
2. What are the characteristics of an ellipse in standard form?
3. How do you identify the lengths of the major and minor axes from the standard form of an ellipse?
4. What is the difference between the standard forms of an ellipse and a circle?
5. How do changes in coefficients affect the orientation of the ellipse?

### Tip:
Always double-check your completed square process, especially the signs and constants, to avoid small errors that can lead to incorrect results.

Learn how to convert ellipse equations to standard form using completing the square method. Identify the center, vertices, and co-vertices of the ellipse based on given equations. Suitable for advanced high school students studying conic sections and algebraic methods.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation