(y/7)^2 + ((x-1)/3)^2 = 1
How do I go about doing this?
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
$\displaystyle \dfrac{(x-3)^2}{9}+\dfrac{y^2}{49}=1 ~\implies~y^2=49-\dfrac{49}9 \cdot (x-3)^2$
Can you take it from here?
- 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
$\displaystyle \pm \frac{7}{3} \sqrt{-x^2+2x+8}$