# Math Help - Rearranging more formulae

1. ## Rearranging more formulae

Hi all,

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

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

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

Here is my attempt for the first one:

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

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

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

And the second one:

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

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

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

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

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

The answers in the back of the book are:

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

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

Could someone please tell me where I am going wrong?

Regards,

evanator

2. Originally Posted by evanator
Hi all,

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

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

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

Here is my attempt for the first one:

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

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

$y = 4\sqrt{1 - (\frac{x}{2})^2}$
If x= 0, $\left(\frac{x}{2}\right)^2+ \left(\frac{y}{4}\right)^2= 1$ becomes $\left(\frac{y}{4}\right)^2= 1$

If y= 4, then $\left(\frac{4}{4}\right)^2= 1^2= 1$ so y= 4 satisfies that.

But if y= -4, then $\left(\frac{-4}{4}\right)^2= (-1)^2= 1$ so y= -4 also satisfies that. Any time you take a square root to solve a quadratic equation, you need both + and -: " $\pm$.

And the second one:

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

$(x + 1)^2 + (y + 1)^2 = 1$
Very good!

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

$y + 1 = \sqrt{1 - (x + 1)^2}$
Again, squaring a positive or negative number can give you the same thing- you need " $\pm$". This should be
$y+ 1= \pm\sqrt{1- (x+ 1)^2}$

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

The answers in the back of the book are:

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

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

Could someone please tell me where I am going wrong?

Regards,

evanator
In both cases you are taking only the "positive" solution to a quadratic.

If $x^2= 4$ then x can be either 2 or -2: $x= \pm 2$.