How do I graph $9 x^{2} + 16 y^{2} - 36 x + 32 y = 92$ $9x^2+16y^2-36x+32y=92$ algebraically?

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Answer 1 · 2014-10-19T21:43:58

By grouping $x$ $x$ 's and $y$ $y$ 's together,

$\Rightarrow (9 x^{2} - 36 x) + (16 y^{2} + 32 y) = 92$ $Rightarrow (9x^2-36x)+(16y^2+32y)=92$

by factoring out 9 and 16,

$\Rightarrow 9 (x^{2} - 4 x) + 16 (y^{2} + 2 y) = 92$ $Rightarrow9(x^2-4x)+16(y^2+2y)=92$

by adding and subtracting 4 and 1 in the parentheses,

$\Rightarrow 9 (x^{2} - 4 x + 4 - 4) + 16 (y^{2} + 2 y + 1 - 1) = 92$ $Rightarrow9(x^2-4x+4-4)+16(y^2+2y+1-1)=92$

by keeping the first three terms in each group,

$\Rightarrow 9 (x^{2} - 4 x + 4) - 36 + 16 (y^{2} + 2 y + 1) - 16 = 92$ $Rightarrow9(x^2-4x+4)-36+16(y^2+2y+1)-16=92$

by completing the squares and adding 36 and 16,

$\Rightarrow 9 {(x - 2)}^{2} + 16 {(y + 1)}^{2} = 144$ $Rightarrow9(x-2)^2+16(y+1)^2=144$

by dividing by 144,

$\Rightarrow \frac{{(x - 2)}^{2}}{16} + \frac{{(y + 1)}^{2}}{9} = 1$ $Rightarrow(x-2)^2/{16}+(y+1)^2/{9}=1$

Hence, we have an equation of an ellipse

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

which is 8 units wide, 6 units tall, and centered at $(2, - 1)$ $(2,-1)$ .

enter image source here

I hope that this was helpful.

How do I graph 9x^2+16y^2-36x+32y=929x2+16y2−36x+32y=929x^2+16y^2-36x+32y=92 algebraically?