How do you find the standard form of $9x^2-16y^2-90x+32y+65=0$ ?

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Answer 1 · 2016-04-15T14:59:54

$(x-5)^2/4^2-(y-1)^2/3^2=1$

$9x^2-90x=9(x-5)^2-225$ .
$-19y^2+32y=-16(y-1)^2+16$ .

So, the given equation becomes
$9(x-5)^2-16(y-1)^2=144$
Divide by 144.
$(x-5)^2/4^2-(y-1)^2/3^2=1$ .

This equation represents a hyper bola with semi-major axis a = 4 and semi-transverse axis b = 3.

The eccentricity of the hyperbola is 1.25, using the relation $b^2=a^2(e^2-1)$

The asymptotes are $y=+-(b/a)x=+-3/4x$ .

How do you find the standard form of 9x^2-16y^2-90x+32y+65=09x^2-16y^2-90x+32y+65=0?