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 ...

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

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

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

What are the vertices of 9x^2 + 16y^2 = 1449x^2 + 16y^2 = 144?