Math Problem Statement

  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

Solution

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 xx and yy terms.

1. Equation: x2+100y2+2x1000y+2401=0x^2 + 100y^2 + 2x - 1000y + 2401 = 0

Step 1: Group the xx and yy terms together. x2+2x+100y21000y+2401=0x^2 + 2x + 100y^2 - 1000y + 2401 = 0

Step 2: Move the constant term to the other side. x2+2x+100y21000y=2401x^2 + 2x + 100y^2 - 1000y = -2401

Step 3: Complete the square for the xx and yy terms.

For xx: x2+2x=(x+1)21x^2 + 2x = (x + 1)^2 - 1

For yy: 100(y210y)=100[(y5)225]=100(y5)22500100(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)21+100(y5)22500=2401(x + 1)^2 - 1 + 100(y - 5)^2 - 2500 = -2401 (x+1)2+100(y5)22501=2401(x + 1)^2 + 100(y - 5)^2 - 2501 = -2401 (x+1)2+100(y5)2=100(x + 1)^2 + 100(y - 5)^2 = 100

Step 5: Divide the entire equation by 100 to get the standard form. (x+1)2100+(y5)21=1\frac{(x + 1)^2}{100} + \frac{(y - 5)^2}{1} = 1

Identifying the center, vertices, and co-vertices:

  • Center: (1,5)(-1, 5)
  • Vertices: For the xx-term (major axis), ±10\pm 10 from the center:
    • (11,5)(-11, 5) and (9,5)(9, 5)
  • Co-vertices: For the yy-term (minor axis), ±1\pm 1 from the center:
    • (1,4)(-1, 4) and (1,6)(-1, 6)

2. Equation: 9x2+25y236x+50y164=09x^2 + 25y^2 - 36x + 50y - 164 = 0

Step 1: Group the xx and yy terms together. 9x236x+25y2+50y=1649x^2 - 36x + 25y^2 + 50y = 164

Step 2: Factor out the coefficients of x2x^2 and y2y^2 terms. 9(x24x)+25(y2+2y)=1649(x^2 - 4x) + 25(y^2 + 2y) = 164

Step 3: Complete the square for the xx and yy terms.

For xx: x24x=(x2)24x^2 - 4x = (x - 2)^2 - 4

For yy: y2+2y=(y+1)21y^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(x2)236+25(y+1)225=1649(x - 2)^2 - 36 + 25(y + 1)^2 - 25 = 164 9(x2)2+25(y+1)2=2259(x - 2)^2 + 25(y + 1)^2 = 225

Step 5: Divide the entire equation by 225 to get the standard form. (x2)225+(y+1)29=1\frac{(x - 2)^2}{25} + \frac{(y + 1)^2}{9} = 1

Identifying the center, vertices, and co-vertices:

  • Center: (2,1)(2, -1)
  • Vertices: For the xx-term (major axis), ±5\pm 5 from the center:
    • (3,1)(-3, -1) and (7,1)(7, -1)
  • Co-vertices: For the yy-term (minor axis), ±3\pm 3 from the center:
    • (2,4)(2, -4) and (2,2)(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.

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Math Problem Analysis

Mathematical Concepts

Conic Sections
Ellipse
Completing the Square

Formulas

Standard form of an ellipse

Theorems

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Suitable Grade Level

Advanced High School