How do I convert the equation $9 x^{2} - y^{2} + 54 x + 10 y + 55 = 0$ $9x^2−y^2+54x+10y+55=0$ to standard form?

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Answer 1 · 2018-08-07T18:24:29

$a = \frac{1}{3}, b = 1, h = - 3, k = 5$ $a = 1/3, b = 1, h = -3, k = 5$

${(3 x + 9)}^{2} - 81 - {(y - 5)}^{2} + 25 = - 55$ $(3x + 9)^2 - 81 - (y - 5)^2 + 25 = - 55$

$9 {(x + 3)}^{2} - {(y - 5)}^{2} = 1$ $9 (x + 3)^2 - (y - 5)^2 = 1$

$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x - h)^2/a^2 - (y-k)^2/b^2 = 1$

How do I convert the equation 9x^2−y^2+54x+10y+55=09x2−y2+54x+10y+55=09x^2−y^2+54x+10y+55=0 to standard form?