How do I convert the equation $- 9 x^{2} + 4 y^{2} + 72 x - 16 y = 164$ $-9x^2 +4y^2 +72x-16y =164$ to standard form?

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Answer 1 · 2014-09-25T02:11:47

We need to complete the square for the $x and x^{2}$ $x and x^2$ terms and then for the $y and y^{2}$ $y and y^2$ terms.

First, reorder the terms

$- 9 x^{2} + 72 x + 4 y^{2} - 16 y = 164$ $-9x^2+72x+4y^2-16y=164$

Factor out a $- 9$ $-9$ from the first 2 terms.
*Remember to change the signs.

Factor out a $4$ $4$ from the last 2 terms.

$- 9 (x^{2} - 8 x) + 4 (y^{2} - 4 y) = 164$ $-9(x^2-8x)+4(y^2-4y)=164$

Work with the $x$ $x$ term.

${(- \frac{8}{2})}^{2} = {(- 4)}^{2} = 16$ $(-8/2)^2=(-4)^2=16$

Remember that we factor out $- 9$ $-9$ so, we have to add $- 9 \cdot 4 = - 36$ $-9*4=-36$ to the right side of the equation.

Work with the $y$ $y$ term.

${(- \frac{4}{2})}^{2} = {(- 2)}^{2} = 4$ $(-4/2)^2=(-2)^2=4$

Remember that we factor out $4$ $4$ so, we have to add $4 \cdot 4 = 16$ $4*4=16$ to the right side of the equation.

$- 9 (x^{2} - 8 x + 16) + 4 (y^{2} - 4 y + 4) = 164 - 144 + 16$ $-9(x^2-8x+16)+4(y^2-4y+4)=164-144+16$

Factor:

$x^{2} - 8 x + 16 \Rightarrow {(x - 4)}^{2} \Rightarrow$ $x^2-8x+16=>(x-4)^2 =>$ Perfect square trinomial

$y^{2} - 4 y + 4 \Rightarrow {(y - 2)}^{2} \Rightarrow$ $y^2-4y+4=>(y-2)^2 =>$ Perfect square trinomial

$- 9 {(x - 4)}^{2} + 4 {(y - 2)}^{2} = 164 - 144 + 16$ $-9(x-4)^2+4(y-2)^2=164-144+16$

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

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

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

How do I convert the equation -9x^2 +4y^2 +72x-16y =164−9x2+4y2+72x−16y=164-9x^2 +4y^2 +72x-16y =164 to standard form?