How would I solve this:

Find a parametric equation for the conic section:

9x^2 - 18x + 6y^2 - 96y + 9 = 0

Thanks!

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- Dec 3rd 2009, 03:22 PMiheartyou119Parametric Equations for Conic Section
How would I solve this:

Find a parametric equation for the conic section:

9x^2 - 18x + 6y^2 - 96y + 9 = 0

Thanks! - Dec 3rd 2009, 04:26 PMbigwave
this is an ellepse

the equation is in the form of

$\displaystyle \frac{\left(x-h\right)^2}{b^2} + \frac{\left(y-k\right)^2}{a^2} = 1$

you will need to complete the square to get it to this form

I was a not exactly sure what you meant by parametric equation

this is just a modified equation from standard form to make is easy to recognize the charactoristics. - Dec 4th 2009, 07:33 AMGrandad
Hello iheartyou119

Welcome to Math Help Forum!As bigwave has said, you'll need to complete the square to write the equation in a form suitable for parametric equations. This form is usually written:$\displaystyle \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$which is indeed an ellipse, centre $\displaystyle (h,k)$. The parametric form of this is:$\displaystyle x = h+a\cos\theta, y = k+b\sin\theta$which works, of course, because $\displaystyle \cos^2\theta + \sin^2\theta = 1$.

Here's the first couple of lines showing you how to complete the square:$\displaystyle 9x^2 - 18x + 6y^2 - 96y + 9 = 0$Can you take it from here and find the values of $\displaystyle a, b, h, k$?

$\displaystyle \Rightarrow 9(x^2-2x) +6(y^2-16y)+9=0$

$\displaystyle \Rightarrow 9\Big((x-1)^2-1\Big)+6\Big((y-8)^2-64\Big)+9=0$

Grandad - Dec 4th 2009, 08:26 AMbigwaveellipse where the foci are paralel to the y axis
I think this an ellipse where the foci are paralel to the y axis

so the $\displaystyle b^2$ would be under the $\displaystyle x^2$

just being cranky - Dec 6th 2009, 12:16 PMiheartyou119still wrong
I have tried this, and I still can't seem to get the right answer. The answer is supposed to be:

x = 1 + 4 cos t, y = 3 + 3 sin t - Dec 6th 2009, 10:09 PMGrandad
Hello iheartyou119The easiest way to check this is to work backwards. These two equations give:

$\displaystyle \cos^2t+\sin^2t =\frac{(x-1)^2}{4^2}+\frac{(y-3)^2}{3^2}=1$

$\displaystyle \Rightarrow 9(x^2-2x+1)+16(y^2-6y+9)=144$

$\displaystyle \Rightarrow 9x^2 -18x+\color{red}16\color{black}y^2-96y+9=0$

So there's the problem: either you copied the question down wrong, or there's a typo in the book.

Grandad - Dec 7th 2009, 12:06 PMiheartyou119wrong equation
Yes, you are correct. The equation is supposed to be:

9x^2 - 18x + 16y^2 - 96y + 9 = 0