How do you write $6x^2 - 24x - 5y^2 - 10y - 11 = 0$ in standard form?

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Answer 1 · 2015-05-21T14:47:37

The standard form of a hyperbola with a transverse axis (all stuff this is, trust me on this) is
$(x-h)^2/(a^2) - (y-k)^2/b^2 = 1$

$6x^2-24x-5y^2-10y-11=0$

move he constant to the left side

$6(x^2-4x) - 5(y^2+2y) = 11$

complete the squares

$6(x^2-4x+4) -5(y^2+2y+1) = 11 +24 -5$

$6(x-2)^2 - 5(y+1)^2 = 30$

Divide by 30 so left side equals 1

$(x-2)^2/(sqrt(5))^2 - (y+1)^2/(sqrt(6))^2 = 1$

How do you write 6x^2 - 24x - 5y^2 - 10y - 11 = 0 6x^2 - 24x - 5y^2 - 10y - 11 = 0 in standard form?