# Thread: Putting the equation of a conic into standard form

1. ## Putting the equation of a conic into standard form

Hi, I have the following exercise...

By completing the square in $\displaystyle x$ and $\displaystyle y$, and so by eliminating the linear terms, put the conic

$\displaystyle 9x^2+4y^2-36x-24y+36=0$

into standard form

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Well, I have completed the square, but how do I get the standard form of this conic?? :-/

$\displaystyle (3x-6)^2+(2y-6)^2=36$

2. Originally Posted by juanma101285
Hi, I have the following exercise...

By completing the square in $\displaystyle x$ and $\displaystyle y$, and so by eliminating the linear terms, put the conic

$\displaystyle 9x^2+4y^2-36x-24y+36=0$

into standard form

___

Well, I have completed the square, but how do I get the standard form of this conic?? :-/

$\displaystyle (3x-6)^2+(2y-6)^2=36$
$\displaystyle [3(x-2)]^2 + [2(y-3)]^2 = 36$

$\displaystyle 9(x-2)^2 + 4(y-3)^2 = 36$

$\displaystyle \dfrac{9(x-2)^2}{36} + \dfrac{4(y-3)^2}{36} = 1$

$\displaystyle \dfrac{(x-2)^2}{2^2} + \dfrac{(y-3)^2}{3^2} = 1$

3. Hello, juanma101285!

$\displaystyle \text{Write the conic: }\:9x^2+4y^2-36x-24y+36\:=\:0\:\text{ in standard form.}$

I recommend that you use this procedure . . .

$\displaystyle \text{Re-arrange terms: }\:9x^2 - 36x + 4y^2 - 24y \;=\;-36$

$\displaystyle \text{Factor: }\quad 9(x^2 - 4x) + 4(y^2 - 6y) \;=\;-36$

$\displaystyle \text{Complete the square:}$

. . . $\displaystyle 9(x^2 - 4x {\bf+ 4 - 4}) + 4(y^2 - 6y {\bf\:+\: 9 - 9}) \;=\;-36$

. . $\displaystyle 9(x^2-4x+4) - 36 + 4(y^2-6y+9) - 36 \;=\;-36$

. . . . . . . . . . . . . . . . .$\displaystyle 9(x-2)^2 + 4(y-3)^2 \;=\;36$

$\displaystyle \text{Divide by 36: }\;\;\dfrac{(x-2)^2}{4} + \dfrac{y-3)^2}{9} \;=\;1$