# What are the vertices of 9x^2 + 16y^2 = 144?

##### 1 Answer
Oct 4, 2014

$9 {x}^{2} + 16 {y}^{2} = 144$

Divide each term by $144.$

$\frac{9 {x}^{2}}{144} + \frac{16 {y}^{2}}{144} = \frac{144}{144}$

Simplify

$\frac{{x}^{2}}{16} + \frac{{y}^{2}}{9} = 1$

The major axis is the x-axis because the largest denominator is under the ${x}^{2}$ term.

The coordinates of the vertices are as follows ...

$\left(\pm a , 0\right)$
$\left(0 , \pm b\right)$

${a}^{2} = 16 \to a = 4$
${b}^{2} = 4 \to b = 2$

$\left(\pm 4 , 0\right)$
$\left(0 , \pm 2\right)$