How do you write the standard form of the hyperbola $-16x^2+9y^2+32x+144y-16=0$ ?

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Answer 1 · 2016-10-17T11:47:18

We can rewrite this as follows

$16x^2-9y^2-32x-144y+16=0$ (Multiply with $-1$ both sides)

$16*(x^2-2x)-9(y^2+16y)+16=0$

$16*(x^2-2x+1)-9(y^2+16+64)+(16-16+576)=0$

$16*(x-1)^2-9*(y+8)^2=-576$

$(y+8)^2/64-(x-1)^2/36=1$

$((y+8)/8)^2-((x-1)/6)^2=1$

Finally the standard form for the hyperbola is

$(y-(-8))^2/8^2-(x-1)^2/6^2=1$

If we plot it on the cartesian plane it is

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How do you write the standard form of the hyperbola -16x^2+9y^2+32x+144y-16=0-16x^2+9y^2+32x+144y-16=0?