Question 433027
put 4x^2-9y^2=-1 in standard form and graph
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Standard form of hyperbola: (x-h)^2/a^2-(y-k)^2/b^2=1 (for horizontal transverse axis)
                                             (y-k)^2/a^2-(x-h)^2/b^2=1 (for vertical transverse axis)
4x^2-9y^2=-1
9y^2-4x^2=1
y^2/(1/9)-x^2/(1/4)=1
This is a hyperbola with center at (0,0) and a vertical transverse axis. (curve opens up and down)
a^2=1/9
a=1/3
b^2=1/4
b=1/2

see graph of given hyperbola below:

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y=+-((1+4x^2)/9)^.5
{{{ graph( 300, 300, -2, 2, -2, 2,((1+4x^2)/9)^.5,-((1+4x^2)/9)^.5) }}}