Math Problem Statement

Convert general form to standard form: 9x^2 + 4y^2 + 72x - 48y + 144 = 0

Solution

Let's work through the conversion of the given equation from general form to standard form step by step. The general form of the equation is:

9x2+4y2+72x48y+144=09x^2 + 4y^2 + 72x - 48y + 144 = 0

Step 1: Group the xx-terms and the yy-terms.

We group the quadratic and linear terms involving xx and yy:

9(x2+8x)+4(y212y)=1449(x^2 + 8x) + 4(y^2 - 12y) = -144

Step 2: Complete the square.

Now, we complete the square for both the xx-terms and the yy-terms.

For the xx-terms:

The expression is x2+8xx^2 + 8x. To complete the square, take half of 8, which is 4, and then square it, giving 16. So:

x2+8x=(x+4)216x^2 + 8x = (x + 4)^2 - 16

For the yy-terms:

The expression is y212yy^2 - 12y. Take half of -12, which is -6, and square it to get 36. So:

y212y=(y6)236y^2 - 12y = (y - 6)^2 - 36

Now substitute these into the equation:

9((x+4)216)+4((y6)236)=1449((x + 4)^2 - 16) + 4((y - 6)^2 - 36) = -144

Step 3: Simplify and move constants to the right-hand side.

Distribute the 9 and 4:

9(x+4)2144+4(y6)2144=1449(x + 4)^2 - 144 + 4(y - 6)^2 - 144 = -144

Combine like terms:

9(x+4)2+4(y6)2288=1449(x + 4)^2 + 4(y - 6)^2 - 288 = -144

Add 288 to both sides:

9(x+4)2+4(y6)2=1449(x + 4)^2 + 4(y - 6)^2 = 144

Step 4: Divide by 144 to get the equation in standard form.

Divide everything by 144:

9(x+4)2144+4(y6)2144=1\frac{9(x + 4)^2}{144} + \frac{4(y - 6)^2}{144} = 1

Simplify the fractions:

(x+4)216+(y6)236=1\frac{(x + 4)^2}{16} + \frac{(y - 6)^2}{36} = 1

Step 5: Identify the values of hh, kk, DD, and EE.

From the standard form:

(xh)2D+(yk)2E=1\frac{(x - h)^2}{D} + \frac{(y - k)^2}{E} = 1

we can compare with our result:

(x+4)216+(y6)236=1\frac{(x + 4)^2}{16} + \frac{(y - 6)^2}{36} = 1

Thus:

  • h=4h = -4
  • k=6k = 6
  • D=16D = 16
  • E=36E = 36

Would you like further details on any of the steps, or do you have any questions?

Related Questions:

  1. How do you complete the square in quadratic expressions?
  2. What is the process for converting a general conic equation to standard form?
  3. What do hh and kk represent in the standard form of an ellipse?
  4. How can you determine if a given conic equation represents a circle, ellipse, or hyperbola?
  5. Why do we divide both sides of the equation by the constant term to standardize the form?

Tip:

Always double-check your completion of the square and constants when converting from general to standard form to avoid mistakes.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Algebra
Conic Sections
Completing the Square
Ellipse

Formulas

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

Theorems

Completing the square

Suitable Grade Level

Grades 10-12