Problem

Find the Center 81x^2+36y^2+972x-648y+2916=0

The question requires you to find the coordinates of the center of an ellipse represented by the given quadratic equation 81x^2 + 36y^2 + 972x - 648y + 2916 = 0. To achieve this, the equation of the ellipse must be rewritten in standard form by completing the squares for both x and y terms, which allows for identifying the center, axes, and orientation of the ellipse.

81x2+36y2+972x648y+2916=0

Answer

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Solution:

Step:1

Convert the given equation into the standard form of an ellipse.

Step:1.1

Subtract 2916 on both sides to isolate the variable terms.

81x2+36y2+972x648y=2916

Step:1.2

Complete the square for the x-terms, 81x2+972x.

Step:1.2.1

Identify coefficients for ax2+bx+c where a=81, b=972, and c=0.

Step:1.2.2

Use the vertex form a(x+d)2+e for a parabola.

Step:1.2.3

Calculate d with d=b2a.

Step:1.2.3.1

Insert a and b into the formula.

d=972281

Step:1.2.3.2

Simplify the fraction by canceling common factors.

Step:1.2.3.2.1

Factor out a 2 from 972.

d=2486281

Step:1.2.3.2.2

Remove the common factor of 2.

d=2486281

Step:1.2.3.2.3

Simplify the fraction.

d=48681

Step:1.2.3.2.4

Reduce the fraction by canceling the common factor of 486 and 81.

d=61 d=6

Step:1.2.4

Determine e with e=cb24a.

Step:1.2.4.1

Plug in c, b, and a.

e=09722481

Step:1.2.4.2

Simplify the expression.

Step:1.2.4.2.1

Perform the arithmetic operations.

e=0944784324 e=2916

Step:1.2.5

Insert a, d, and e into the vertex form.

81(x+6)22916

Step:1.3

Replace 81x2+972x with 81(x+6)22916 in the original equation.

81(x+6)2+36y2648y=2916

Step:1.4

Cancel out 2916 on both sides.

81(x+6)2+36y2648y=0

Step:1.5

Complete the square for the y-terms, 36y2648y.

Step:1.5.1

Identify coefficients for ay2+by+c where a=36, b=648, and c=0.

Step:1.5.2

Use the vertex form a(y+d)2+e for a parabola.

Step:1.5.3

Calculate d with d=b2a.

Step:1.5.3.1

Insert a and b into the formula.

d=648236

Step:1.5.3.2

Simplify the fraction by canceling common factors.

Step:1.5.3.2.1

Factor out a 2 from 648.

d=2324236

Step:1.5.3.2.2

Remove the common factor of 2.

d=2324236

Step:1.5.3.2.3

Simplify the fraction.

d=32436

Step:1.5.3.2.4

Reduce the fraction by canceling the common factor of 324 and 36.

d=91 d=9

Step:1.5.4

Determine e with e=cb24a.

Step:1.5.4.1

Plug in c, b, and a.

e=0(648)2436

Step:1.5.4.2

Simplify the expression.

e=2916

Step:1.5.5

Insert a, d, and e into the vertex form.

36(y9)22916

Step:1.6

Replace 36y2648y with 36(y9)22916 in the original equation.

81(x+6)2+36(y9)2=0

Step:1.7

Eliminate 2916 by adding 2916 to both sides.

81(x+6)2+36(y9)2=2916

Step:1.8

Simplify the equation to get the standard form.

(x+6)236+(y9)281=1

Step:2

Recognize that the equation is now in the standard form of an ellipse.

Step:3

Identify the center and axes lengths from the standard form.

a=9,b=6,h=6,k=9

Step:4

The center of the ellipse is given by the coordinates (h,k).

Center=(6,9)

Knowledge Notes:

To solve for the center of an ellipse given in the general quadratic form, one must first rewrite the equation in the standard form of an ellipse. The standard form for an ellipse is:

(xh)2a2+(yk)2b2=1 where (h,k) is the center of the ellipse, a is the semi-major axis, and b is the semi-minor axis.

The process involves:

  1. Rearranging the equation to group the x and y terms.

  2. Completing the square for both x and y terms to isolate them on one side.

  3. Dividing through by the constant term to normalize the equation to equal 1.

  4. Comparing the resulting equation to the standard form to identify the center and axes lengths.

The vertex form of a parabola is used to complete the square, which is a(x+d)2+e for x-terms and a(y+d)2+e for y-terms. The value of d is found using the formula d=b2a, and e is found using e=cb24a, where a, b, and c are coefficients from the general quadratic form ax2+bx+c.

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