# Thread: help solving for y

1. ## help solving for y

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

How do I go about doing this?

2. Originally Posted by clambottomjewels
(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?

3. Originally Posted by clambottomjewels
(y/7)^2 + ((x-1)/3)^2 = 1

How do I go about doing this?
• 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

$\displaystyle \pm \frac{7}{3} \sqrt{-x^2+2x+8}$