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

Question

anonymous · Answer

The hyperbola has the simplified equation

(4x-8)²/20² - (5y+5)²/ 20² =1

4²(x-2)²/20² - 5²(y+1)²/ 20² =1
(x-2)²/5² -(y+1)²/4² =1
in new coordinates
X=x-2
Y=y+1
we have

X²/5² - Y²/4² =1
The foci are (-c,0), (c,0)
where c²=a²+b²
a²=25
b²=16
c²=41

c=√41
F1=(-√41 ,0)
F2=(√41,0)

In original coordinates
f1= (2-√41 , -1)
f2=(2+√41 , -1)

anonymous · Answer

wow that was quick and thorough!

anonymous · Answer

YOU"RE AWSOME