Outcome E14(adv): translate between different forms of equations of circles and ellipses
Originator: Alison
Formulas:
Standard form (circle): (x-h)2 + (y-k)2 = r2
General form (circle): Ax2 + Ay2 + Dx + Ey + F = 0
Transformational form (circle): [1/r(x-h)]2 +[1/r(y-k)]2 = 1
Transformational form (ellipse): [1/a(x-h)]2 +[1/b(y-k)]2 = 1
General form (ellipse): Ax2 + By2 + Dx + Ey + F = 0
Explanation:
General -> Standard form (circles): First, gather like terms and move the F term to the other side of the equals sign. Factor out the A value and put like terms in brackets. Complete the square of the like terms inside the brackets and add each value to the other side of the equation. Simplify. Divide both sides by the coefficient A.
Standard -> Transformational (circles): Make the right side of the equation equal to 1 by dividing both sides by the number that is there. Square root the denominators on the left side, put square brackets around each of the 2 terms and square both terms.
Transformational -> Standard (circles): Romove the square brackets from both terms on the left side of the equation. Get rid of the fractional coefficients by multiplying both sides by the denominator.
General -> Transformational (ellipse): Gather like terms and move the F value to the other side of the equation. Factor the A and B values out of the like terms and complete the square inside the brackets. Add this value to the other side aswell. Simplify. Make the right side of the equation equal to 1 by dividing both sides by the number that is there. Place square brackets around both terms seperatly and move the 2 to the outside of them.
Transformational -> General (ellipse): Remove square brackets. Multiply all terms in the equation by the denominators outside the brackets. Multiply out the brackets by their coefficients and gather like terms. Make the right side of the equation equal to 0 by subtracting whatever is there from both sides.
Examples:
General to standard form (circle):
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Start
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2x2 +2y2 +12x +8y – 6 = 0
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Step 1
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2x2 + 12x + 2y2 + 8y = 6
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Step 2
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2(x2 + 6x) + 2(y2 + 4y) = 6
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Step 3
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2(x2 + 6x + 9) + 2(y2 + 4y + 4) = 6 + 18 + 8
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Step 4
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2(x + 3) 2 + 2(y + 2) 2 =32
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Step 5
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(x + 3) 2 + (y + 2) 2 =16
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Standard to transformational form (circle):
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Start
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(x + 3) 2 + (y + 2) 2 =16
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Step 1
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1/16(x + 3) 2 + 1/16(y + 2) 2 =16/16
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Step 2
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[1/4(x + 3)] 2+ [1/4(y + 2)] 2=1
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Transformational to standard form (circle):
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Start
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[1/4(x + 3)] 2 + [1/4(y + 2)] 2 =1
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Step 1
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1/16(x + 3) 2 + 1/16(y + 2) 2 =1
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Step 2
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(x + 3) 2 + (y + 2) 2 =16
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General to Transformational form (ellipse):
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Start
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9x2 + 4y2 + 18x – 16y – 11 = 0
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Step 1
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9x2 +18x + 4y2 – 16 y = 11
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Step 2
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9(x2 + 2x) + 4(y2 - 4x) = 11
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Step 3
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9(x2 + 2x + 1) + 4(y2 - 4x + 4) = 11 + 9 + 16
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Step 4
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9(x + 1) 2 + 4(y – 2) 2 = 36
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Step 5
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9/36(x + 1) 2 + 4/36(y – 2) 2 = 36/36
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Step 6
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[1/2(x + 1)]2 + [1/3(y – 2)] 2 = 1
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Transformational to general form (ellipse):
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Start
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[1/2(x + 1)]2 + [1/3(y – 2)] 2 = 1
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Step 1
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1/4(x + 1)2 + 1/9(y – 2) 2 = 1
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Step 2
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9(x + 1) 2 + 4(y – 2) 2 = 9 * 4
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Step 3
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9(x2 + 2x + 1) + 4(y2 - 4x + 4) = 36
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Step 4
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9x2 +18x + 9 + 4y2 – 16y + 16 = 36
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Step 5
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9x2 +18x + 4y2 – 16 y +25 = 36
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Step 6
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9x2 + 4y2 + 18x – 16y – 11 = 0
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Sample Question:
Change the following equation into transformational form and identify whether it is a circle or an ellpise:
(a) (x + 1) 2 + (x – 4) 2 = 9
(b) 4x2 + 16y2 - 12x + 32y + 16 = 0
Solutions:
(a) (x + 1) 2 + (x – 4) 2 = 9
1/9(x + 1) 2 + 1/9(x – 4) 2 = 9/9
[1/3(x + 1)] 2 + [1/3(x – 4)] 2 = 9 [I think you mean 1 here...]
This equation is a circle because the 'r' vaules are equal.
(b) 4x2 + 16y2 - 12x [should this be -16x?] + 32y + 16 = 0
4x2 - 16x + 16y2 + 32y = -16
4(x2 - 4x) + 16(y2 + 2y) = -16
4(x2 - 4x + 4) + 16(y2 + 2y + 1) = -16 +16 +16
4(x - 2)2 + 16(y + 1)2 = 16
4/16(x - 2)2 + 16/16(y + 1)2 = 16/16
[1/4(x - 2)]2 + [1(y + 1)]2 = 1
This equation is an ellipse because the denominators of the fractions are different.
Comments (1)
Anonymous said
at 7:26 pm on Jun 14, 2007
Nice job... just some slight problems with your sample questions.
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