help solving for y

**clambottomjewels** · November 1st 2009, 10:16 AM

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

How do I go about doing this?

**earboth** · November 1st 2009, 10:23 AM

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:

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

So what exactly do you want to do with it?

EDIT: OK, finally I read the headline

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

Can you take it from here?

**e^(i*pi)** · November 1st 2009, 10:25 AM

Originally Posted by clambottomjewels

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

How do I go about doing this?

Take $\left(\frac{1}{3}(x-1)\right)^2$ from both sides
Take the square root of both sides (and don't forget the $\pm$
Multiply both sides by 7

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

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

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