document.write( "Question 620155: 64x^2+16y^2+320x-192y+960=0
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Algebra.Com's Answer #389963 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
 
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document.write( "Hi,
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document.write( "64x^2+16y^2+320x-192y+960=0
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document.write( " 64(x +5/2)^2 - 64(25/4)  +  16(y+6)^2 - 16(36) +960 = 0
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document.write( " 64(x +5/2)^2 +  16(y+6)^2 = 16
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document.write( "  4(x+5/2)^2  + (y+6)^2 =1  Yes, good Work.
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document.write( "   \"%28x%2B+5%2F2%29%5E2%2F%281%2F2%29%5E2+%2B+%28y%2B6%29%5E2%2F1%5E2+=+1\" This format gives the standard form
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document.write( " C(-5/2,-6)   V(-3,-6)   &    V(-2,6)  and   V(-5/2,-7)  &  V(-5/2,-5)
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document.write( "sqrt(1-1/4) = sqrt(3/4) = ħ sqrt(3)/2  F(-5/2, -6ħsqrt(3)/2)  
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document.write( " Standard Form of an Equation of an Ellipse is \"%28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1+\" where Pt(h,k) is the center. (a positioned to correspond with major axis)
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document.write( " a and b  are the respective vertices distances from center and ħ \"sqrt%28a%5E2-b%5E2%29\"are the foci distances from center: a > b
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