subtract the first from the second:
-3xy + y^2 = 12
x = (12 - y^2)/(-3y) or (y^2 - 12)/(3y)
sub into the first:
(y^2 - 12)^2 / 9y^2 + y^2= 13
(y^2 - 12)^2 + 9y^4 = 117y^2
y^4 - 24y^2 + 144 + 9y^4 - 117y^2 = 0
10y^4 - 141y^2 + 144 = 0
y^2 = (141 ± √14121)/20
y^2 = appr 12.99 or appr 1.1084
y = appr ± 3.604 or appr ±1.0528
Your turn:
Sub those our two y^2 values into
x^2 + y^2 = 13
and find the matching x values.
You should get 4 points, since we are intersecting a circle with a hyperbola
Help. solve simultaneously
x^2+y^2=13 and
x^2-3xy+2y^2=25
1 answer
1 answer