How do you find the standard form given $9 x^{2} - 4 y^{2} - 36 x - 24 y - 36 = 0$ $9x^2-4y^2-36x-24y-36=0$ ?

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Answer 1 · 2018-05-03T12:38:02

$\frac{{(x - 2)}^{2}}{2^{2}} - \frac{{(y + 3)}^{2}}{3^{2}} = 1$ $( x -2 )^2/ 2^2 - (y +3)^2/3^2 = 1$ , equation of a hyperbola .

$9 x^{2} - 4 y^{2} - 36 x - 24 y - 36 = 0$ $9 x^2 -4 y^2 -36 x -24 y -36=0$ or

$9 (x^{2} - 4 x) - 4 (y^{2} + 6 y) = 36$ $9( x^2 -4 x ) -4 (y^2 +6 y) = 36$ or

$9 (x^{2} - 4 x + 4) - 4 (y^{2} + 6 y + 9) = 36$ $9( x^2 -4 x +4 ) -4 (y^2 +6 y +9) = 36$ or

$9 {(x - 2)}^{2} - 4 {(y + 3)}^{2} = 36$ $9( x -2 )^2 - 4 (y +3)^2 = 36$ or

$\frac{9 {(x - 2)}^{2}}{36} - \frac{4 {(y + 3)}^{2}}{36} = 1$ $(9( x -2 )^2)/36 - (4 (y +3)^2)/36 = 1$ or

$\frac{{(x - 2)}^{2}}{4} - \frac{{(y + 3)}^{2}}{9} = 1$ $( x -2 )^2/4 - (y +3)^2/9 = 1$ or

$\frac{{(x - 2)}^{2}}{2^{2}} - \frac{{(y + 3)}^{2}}{3^{2}} = 1$ $( x -2 )^2/ 2^2 - (y +3)^2/3^2 = 1$

This is standard form of the equation of a hyperbola with center

at $(2, - 3)$ $( 2,-3)$

graph{9 x^2-4 y^2-36 x-24 y=36 [-80, 80, -40, 40]} [Ans]

How do you find the standard form given 9x2−4y2−36x−24y−36=09x^2-4y^2-36x-24y-36=0?