Math Problem Statement

Express the equation (x-6)^2/16 + y^2/12 = 1 to its general form and transform the equation 3x^2 + 2y^2 - 6x - 12y + 15 = 0 to standard form.

Solution

Let's go through the problem step by step.

Problem 9: Express the equation

(x6)216+y212=1\frac{(x-6)^2}{16} + \frac{y^2}{12} = 1 to its general form.

This equation is the standard form of an ellipse. To convert it to the general quadratic form Ax2+By2+Cx+Dy+E=0Ax^2 + By^2 + Cx + Dy + E = 0, we'll follow these steps:

  1. Clear the denominators by multiplying through by the least common denominator (LCD). The denominators are 16 and 12, so the LCD is 48. 48×((x6)216+y212=1)48 \times \left( \frac{(x-6)^2}{16} + \frac{y^2}{12} = 1 \right) This simplifies to: 3(x6)2+4y2=483(x-6)^2 + 4y^2 = 48

  2. Expand the squares: 3(x212x+36)+4y2=483(x^2 - 12x + 36) + 4y^2 = 48 3x236x+108+4y2=483x^2 - 36x + 108 + 4y^2 = 48

  3. Move all terms to one side of the equation: 3x2+4y236x+10848=03x^2 + 4y^2 - 36x + 108 - 48 = 0 Simplifying the constants: 3x2+4y236x+60=03x^2 + 4y^2 - 36x + 60 = 0

This matches option (a): 3x2+4y236x+60=0\boxed{3x^2 + 4y^2 - 36x + 60 = 0}

Problem 10: Transform the equation

3x2+2y26x12y+15=03x^2 + 2y^2 - 6x - 12y + 15 = 0 to its standard form.

This is the equation of an ellipse, and we need to complete the square for both xx and yy.

  1. Group the xx-terms and the yy-terms: 3(x22x)+2(y26y)=153(x^2 - 2x) + 2(y^2 - 6y) = -15

  2. Complete the square: For the xx-terms: x22xx^2 - 2x. Take half of 2-2, which is 1-1, and square it to get 1. Add and subtract inside the parentheses: 3((x1)21)3((x - 1)^2 - 1)

    For the yy-terms: y26yy^2 - 6y. Take half of 6-6, which is 3-3, and square it to get 9. Add and subtract inside the parentheses: 2((y3)29)2((y - 3)^2 - 9)

  3. Substitute back into the equation: 3((x1)21)+2((y3)29)=153((x - 1)^2 - 1) + 2((y - 3)^2 - 9) = -15 Simplify: 3(x1)23+2(y3)218=153(x - 1)^2 - 3 + 2(y - 3)^2 - 18 = -15 Combine constants: 3(x1)2+2(y3)221=153(x - 1)^2 + 2(y - 3)^2 - 21 = -15 3(x1)2+2(y3)2=63(x - 1)^2 + 2(y - 3)^2 = 6

  4. Divide by 6 to get the standard form: (x1)22+(y3)23=1\frac{(x - 1)^2}{2} + \frac{(y - 3)^2}{3} = 1

This matches option (a): (x1)22+(y3)23=1\boxed{\frac{(x - 1)^2}{2} + \frac{(y - 3)^2}{3} = 1}

Would you like further details or clarifications? Here are five related questions you might consider:

  1. What is the general form of a conic section?
  2. How do you identify whether a conic section represents an ellipse, hyperbola, or circle?
  3. Can you explain the difference between completing the square and expanding binomials?
  4. How do you find the center, vertices, and foci of an ellipse from its standard form?
  5. How would the process change if the equation represented a hyperbola?

Tip: When completing the square, always double-check the constants added to ensure correct balancing on both sides of the equation.

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Math Problem Analysis

Mathematical Concepts

Algebra
Conic Sections
Ellipses
Standard Form of Equations

Formulas

General form of a conic section: Ax^2 + By^2 + Cx + Dy + E = 0
Standard form of an ellipse: (x-h)^2/a^2 + (y-k)^2/b^2 = 1

Theorems

Completing the square
Properties of ellipses

Suitable Grade Level

Grades 10-12