(y/7)^2 + ((x-1)/3)^2 = 1

How do I go about doing this?

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- Nov 1st 2009, 10:16 AMclambottomjewelshelp solving for y
(y/7)^2 + ((x-1)/3)^2 = 1

How do I go about doing this? - Nov 1st 2009, 10:23 AMearboth
This is the equation of an ellipse:

$\displaystyle \left(\dfrac y7\right)^2+\left(\dfrac{x-3}3\right)^2=1~\implies~\dfrac{(x-3)^2}{9}+\dfrac{y^2}{49}=1$

So what exactly do you want to do with it?

EDIT: OK, finally I read the headline (Blush)

$\displaystyle \dfrac{(x-3)^2}{9}+\dfrac{y^2}{49}=1 ~\implies~y^2=49-\dfrac{49}9 \cdot (x-3)^2$

Can you take it from here? - Nov 1st 2009, 10:25 AMe^(i*pi)
- Take $\displaystyle \left(\frac{1}{3}(x-1)\right)^2$ from both sides
- Take the square root of both sides (and don't forget the $\displaystyle \pm$
- Multiply both sides by 7

You should get an answer of $\displaystyle y = \pm7\sqrt{1-\left(\frac{1}{3}(x-1)\right)^2}$. This is equivalent to

$\displaystyle \pm \frac{7}{3} \sqrt{-x^2+2x+8}$ - Take $\displaystyle \left(\frac{1}{3}(x-1)\right)^2$ from both sides