Find the points where these quadratic relations intersect. 16x^2-5y^2=64, 16x^2+25y^2-96x=256

Answer 1

just substitute:

16x^2 - 96x + 5(16x^2-64) = 256
16x^2 - 96x + 80x^2 - 320 = 256
96x^2 - 96x - 576 = 0
x^2 - x - 6 = 0
(x+2)(x-3) = 0

Now you have x, get y.

see the solution at

http://www.wolframalpha.com/input/?i=solve+16x^2-5y^2%3D64%2C+16x^2%2B25y^2-96x%3D256

Answer 2

subtract them :
30y^2 - 96x = 192
5y^2 = 32 + 16x

back into the first:
16x^2 - (32+16x) = 64
16x^2 -16x - 96 = 0
x^2 - x - 6 = 0
(x-3)(x+2) = 0
x = 3 or x = -2

if x = 3,
5y^2 = 32+48 = 80
y^2 = 16
y = ± 4

if x = -2
5y^2 = 32-32 = 0
y = 0

so we have 3 points:
(-2,0) , (3,4) and (3,-4)

verification:
http://www.wolframalpha.com/input/?i=solve++16x%5E2-5y%5E2%3D64%2C+16x%5E2%2B25y%5E2-96x%3D256