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how do you solve this equation and determine the conic section: 81x^2 + 4y^2 -324 = 0

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how do you solve this equation and determine the conic section: 81x^2 + 4y^2 -324 = 0

asked Nov 30, 2013 in ALGEBRA 2 by angel12 Scholar

3 Answers

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Given equation is 81x^2+4y^2-324 = 0

Divide to each side by 324.

81x^2/324+4y^2/324 -324/324 = 0

(x^2/4+y^2/81)-1 = 0

Add 1 to each side.

(x^2/4+y^2/81)-1+1 = 0+1

x^2/2^2+y^2/9^2 = 1

(x/2)^2+ (y/9)^2 = 1

Compare the ellipse equation (x/a)^2+(y/b)^2 = 1

Here a = 2,b = 9.

Given equation is ellipse equation.

answered Nov 30, 2013 by william Mentor
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The diophantine equation image

The term Diophantine Equation means that the solutions (x , y ) should be integer numbers.

Compare the above equation image.

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So the above equation is elliptical.

Elliptical case

Since the ellipse is a closed figure, the number of solutions will be finite.

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answered Jun 6, 2014 by david Expert
edited Jun 6, 2014 by david
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Continuous...

For any value of x  there will be two values of y  except at the left and right extremes of the ellipse. In this case there will be only one value of y . To determine the location of the left and right extremes we should equal the square root to zero, so the previous expression returns only one value of y .

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The values of x  should be between the roots of image

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Substitute the value of x  in  image

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Substitute = 0 in  image

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Solution

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image.

answered Jun 6, 2014 by david Expert
edited Jun 6, 2014 by david

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