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Convert the standard equation of a hyperbola `2(x-2)^2/37 - (y-5)^2/37 =1` to general...

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ssdixon | Student, Undergraduate | (Level 1) Salutatorian

Posted June 7, 2012 at 4:37 AM via web

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Convert the standard equation of a hyperbola `2(x-2)^2/37 - (y-5)^2/37 =1` to general form Ax^2+Cy^2 - 8x+ Ey +F=0

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted June 7, 2012 at 4:42 AM (Answer #1)

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The equation of the hyperbola `2(x-2)^2/37 - (y-5)^2/37 =1` has to be converted to the general form.

`2(x-2)^2/37 - (y-5)^2/37 =1`

=> 2(x-2)^2 - (y-5)^2 =37

=> 2(x^2 - 4x + 4) - (y^2 - 10y + 25) = 37

=> 2x^2 - 8x + 8 - y^2 + 10y - 25 = 37

=> 2x^2 - y^2 - 8x + 10y + 8 - 25 - 37 = 0

=> 2x^2 - y^2 - 8x + 10y - 54 = 0

The equation of the hyperbola in the general form is 2x^2 - y^2 - 8x + 10y - 54 = 0

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tuhinzia | Student, Undergraduate | (Level 1) Salutatorian

Posted June 7, 2012 at 8:29 AM (Answer #2)

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i have the same ans.

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