# Reduce the conic to standard form

• May 7th 2010, 04:46 AM
aspotonthewall
Reduce the conic to standard form
I have a conic equation

$36y^2+504y-4x^2+48x+1476=0$

Which I am required to write in one of the following formats:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

$y=ax^2$

$y=ay^2$

I've been looking online for a guide on how to do this, but with little success so far. Perhaps any of you can help. Thanks :)
• May 7th 2010, 05:03 AM
skeeter
Quote:

Originally Posted by aspotonthewall
I have a conic equation

$36y^2+504y-4x^2+48x+1476=0$

Which I am required to write in one of the following formats:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

$y=ax^2$

$y=ay^2$

I've been looking online for a guide on how to do this, but with little success so far. Perhaps any of you can help. Thanks :)

start by dividing every term by 4 ...

$9y^2+126y-x^2+12x+369=0$

start the process of completing the squares ...

$9(y^2+14y) - (x^2-12x) = -369$

$9(y^2+14y+49) - (x^2-12x+36) = -369 +9(49) - 36$

$
9(y+7)^2 - (x-6)^2 = 36
$

divide every term by 36 ...

$
\frac{(y+7)^2}{4} - \frac{(x-6)^2}{36} = 1
$

$
\frac{(y+7)^2}{2^2} - \frac{(x-6)^2}{6^2} = 1
$
• May 7th 2010, 05:09 AM
tonio
Quote:

Originally Posted by aspotonthewall
I have a conic equation

$36y^2+504y-4x^2+48x+1476=0$

Which I am required to write in one of the following formats:

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

$\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$

$y=ax^2$

$y=ay^2$

I've been looking online for a guide on how to do this, but with little success so far. Perhaps any of you can help. Thanks :)

Complete the square: if $t$ is a variable/unknown, we have that $At^2\pm Bt=A\left(t\pm\frac{B}{2A}\right)^2-\frac{B^2}{4A}\,,\,\,A\neq 0$

Do this with the variables $x,y$ and then do some algebraic order...BTW, you should get a north-south hyperbola with semiaxis 2 and 6.

Tonio