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

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Answer 1 · 2014-10-04T21:40:48

#9x^2+16y^2=144#

Divide each term by #144.#

#(9x^2)/144+(16y^2)/144=144/144#

Simplify

#(x^2)/16+(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 ...

#(+-a,0)#
#(0,+-b)#

#a^2=16 -> a=4#
#b^2=4 -> b=2#

#(+-4,0)#
#(0,+-2)#