# How do you write 2(x+4)^2 + 3(y-1)^2 = 24 in standard form?

May 3, 2016

${\left(x - \left(- 4\right)\right)}^{2} / {\left(\sqrt{12}\right)}^{2} + {\left(y - 1\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$

#### Explanation:

The standard form for an ellipse is
$\textcolor{w h i t e}{\text{XXX}} {\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$

Given
$\textcolor{w h i t e}{\text{XXX}} 2 {\left(x + 4\right)}^{2} + 3 {\left(y - 1\right)}^{2} = 24$

$\Rightarrow$
$\textcolor{w h i t e}{\text{XXX}} {\left(x + 4\right)}^{2} / 12 + {\left(y - 1\right)}^{2} / 8 = 1$

$\textcolor{w h i t e}{\text{XXX}} {\left(x - \left(- 4\right)\right)}^{2} / {\left(\sqrt{12}\right)}^{2} + {\left(y - 1\right)}^{2} / {\left(\sqrt{8}\right)}^{2} = 1$

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

If you like you could replace ${\left(\sqrt{12}\right)}^{2}$ with ${\left(2 \sqrt{3}\right)}^{2}$
and ${\left(\sqrt{8}\right)}^{2}$ with ${\left(2 \sqrt{2}\right)}^{2}$
...but this really doesn't make the form any simpler.