Hi all,

A problem I was working on required me to rearrange these two equations to find y in terms of x:

$\displaystyle (\frac{x}{2})^2 + (\frac{y}{4})^2 = 1$

$\displaystyle x^2 + y^2 + 2x + 2y + 1 = 0$

Here is my attempt for the first one:

$\displaystyle (\frac{y}{4})^2 = 1 - (\frac{x}{2})^2$

$\displaystyle \frac{y}{4} = \sqrt{1 - (\frac{x}{2})^2}$

$\displaystyle y = 4\sqrt{1 - (\frac{x}{2})^2}$

And the second one:

$\displaystyle x^2 + 2x + y^2 + 2y = -1$

$\displaystyle (x + 1)^2 + (y + 1)^2 = 1$

$\displaystyle (y + 1)^2 = 1 - (x + 1)^2$

$\displaystyle y + 1 = \sqrt{1 - (x + 1)^2}$

$\displaystyle y = \sqrt{1 - (x + 1)^2} - 1$

The answers in the back of the book are:

$\displaystyle y = \pm 4\sqrt{1 - \frac{x^2}{3}}$

$\displaystyle y = \pm \sqrt{1 - (1 - x)^2} - 1$

Could someone please tell me where I am going wrong?

Regards,

evanator