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 · 2017-02-15T06:52:46

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

Let's do some rearrangement by completing the squares

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

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

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

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

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

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

This is the equation of an ellipse.

graph{9x^2-36x+4y^2+24y+36=0 [-11.41, 11.09, -11.1, 0.15]}

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