# Thread: Parametric Equations for Conic Section

1. ## Parametric 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!

2. this is an ellepse

the equation is in the form of

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

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.

3. Hello iheartyou119

Welcome to Math Help Forum!
Originally Posted by iheartyou119
How would I solve this:

Find a parametric equation for the conic section:
9x^2 - 18x + 6y^2 - 96y + 9 = 0

Thanks!
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:
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
which is indeed an ellipse, centre $(h,k)$. The parametric form of this is:
$x = h+a\cos\theta, y = k+b\sin\theta$
which works, of course, because $\cos^2\theta + \sin^2\theta = 1$.

Here's the first couple of lines showing you how to complete the square:
$9x^2 - 18x + 6y^2 - 96y + 9 = 0$

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

$\Rightarrow 9\Big((x-1)^2-1\Big)+6\Big((y-8)^2-64\Big)+9=0$
Can you take it from here and find the values of $a, b, h, k$?

4. ## ellipse 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 $b^2$ would be under the $x^2$

just being cranky

5. ## still 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

6. Hello iheartyou119
Originally Posted by iheartyou119
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
The easiest way to check this is to work backwards. These two equations give:

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

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

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

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

7. ## wrong equation

Yes, you are correct. The equation is supposed to be:

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