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

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How do you write $6 x^{2} - 24 x - 5 y^{2} - 10 y - 11 = 0$ $6x^2 - 24x - 5y^2 - 10y - 11 = 0$ 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
$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x-h)^2/(a^2) - (y-k)^2/b^2 = 1$

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

move he constant to the left side

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

complete the squares

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

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

Divide by 30 so left side equals 1

$\frac{{(x - 2)}^{2}}{{(\sqrt{5})}^{2}} - \frac{{(y + 1)}^{2}}{{(\sqrt{6})}^{2}} = 1$ $(x-2)^2/(sqrt(5))^2 - (y+1)^2/(sqrt(6))^2 = 1$

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

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How do you write 6x2−24x−5y2−10y−11=0 6x^2 - 24x - 5y^2 - 10y - 11 = 0 in standard form?

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