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

1 Answer
Jun 9, 2016

y=-1/36x^2+9/4 which is a parabola with a vertical axis of symmetry and a vertex at (0,9/4)

Explanation:

(x^2)/81+4y/9=1

rArrcolor(white)("XX")4y/9=1-x^2/81

rArrcolor(white)("XX")y=9/4(1-x^2/81)

rArrcolor(white)("XX")y=-1/36x^2+9/4

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