How do you find the standard form of #3x^2/12+ 5y^2/500 = 1# and what kind of a conic is it?

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Answer 1 · 2016-01-04T23:38:57

An ellipse.

Simplify the fractions.

#x^2/4+y^2/100=1#

This fits the mold for an ellipse, which has the general form

#(y-k)^2/a^2+(x-h)^2/a^2=1#

However, the terms at first are switched since #a>b# for all ellipses, here denoting a vertically oriented ellipse.

Thus, the actual general form is

#y^2/100+x^2/4=1#

Here, #h=0,k=0,a=10,b=2#.

graph{y^2/100+x^2/4=1 [-22.81, 22.8, -11.4, 11.41]}