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:$\displaystyle \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$
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:$\displaystyle 9x^2 - 18x + 6y^2 - 96y + 9 = 0$
$\displaystyle \Rightarrow 9(x^2-2x) +6(y^2-16y)+9=0$
$\displaystyle \Rightarrow 9\Big((x-1)^2-1\Big)+6\Big((y-8)^2-64\Big)+9=0$
Can you take it from here and find the values of $\displaystyle a, b, h, k$?
Grandad