What are the foci of the hyperbola given by the equation 16x2 - 25y2 - 64x - 50y - 361 = 0?

Question

anonymous · Answer

so isolate your x's and y's first so that you get....

$$(16x ^{2}-64x) - (25y ^{2}+50y) = 361$$

then you do some factoring out so that you get....

$$(16(x ^{2}-4x)) - (25(y ^{2}+2y)) = 361$$

now you do the squared rule thing to get...

$$(16(x ^{2}-4x+4-4)) - (25(y ^{2}+2y+1-1)) = 361$$

$$16(x-2)^{2}-64 - 25(y+1)^{2}+25 = 361$$

then you just consolidate the x's and y's again....

$$16(x-2)^{2} - 25(y+1)^{2} = 400$$

so now your center is (2, -1)

now divide by 400 on each side to get....

$$\frac{ (x-2)^{2} }{ 25 } - \frac{ (y+1)^{2} }{ 16 } = 1$$

from there you should be able to get your foci