document.write( "Question 658628: Find the verticces for the hyperbola, -x^2+4y^2-6x-48y+139=0, and write your answer in this form: (x1,y1),(x2,y2). \n" ); document.write( "

Algebra.Com's Answer #410355 by ewatrrr(24785) $\"\"$ $\"About$

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document.write( "Hi,
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document.write( "-x^2+4y^2-6x-48y+139=0
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document.write( "x^2-4y^2+6x+48y =139
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document.write( " (x+3)^2 - 4(y-6)^2 = 139 + 9 - 144
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document.write( " (x+3)^2 - 4(y-6)^2 = 4
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document.write( " C(-3,6), V(-5,6) and (-1,6)
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document.write( "Standard Form of an Equation of an Hyperbola opening right and  left is:
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document.write( "   with C(h,k) and vertices 'a' units right and left of center,   
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