# How do you find the standard form of x^2/81 + 4y/9 = 1 and what kind of a conic is it?

Jun 9, 2016

$y = - \frac{1}{36} {x}^{2} + \frac{9}{4}$ which is a parabola with a vertical axis of symmetry and a vertex at $\left(0 , \frac{9}{4}\right)$

#### Explanation:

$\frac{{x}^{2}}{81} + 4 \frac{y}{9} = 1$

$\Rightarrow \textcolor{w h i t e}{\text{XX}} 4 \frac{y}{9} = 1 - {x}^{2} / 81$

$\Rightarrow \textcolor{w h i t e}{\text{XX}} y = \frac{9}{4} \left(1 - {x}^{2} / 81\right)$

$\Rightarrow \textcolor{w h i t e}{\text{XX}} y = - \frac{1}{36} {x}^{2} + \frac{9}{4}$

graph{x^2/81+4y/9=1 [-16.03, 16.01, -8, 8.03]}