Intersection of a hyperbola with an ellipse, cool
2 x^2 - 5 y^2 = 1
2 x^2 + 10y^2 = 136
--------------------subtract
-15 y^2 = -135
y^2 = 9 so y = +/-3
x^2 = 68-45 = 23 so x = +/-sqrt(23)
(sqrt 23 , 3)
(sqrt 23 , -3)
(-sqrt 23 , 3 )
(-sqrt 23 , -3)
Solve the system algebraically
2x^2-5y^2=1
x^2+5y^2=68
Please show work!!!
2 answers
How about just adding them as they sit
3x^2 = 69
x^2 = 23
x = ±√23
plug x^2 = 23 back into the 2nd
23 + 5y^2 = 68
5y^2 = 45
y^2 = 9
y = ±3
so 4 solutions:
(√23,3) , (√23, -3) , (-√23 , 3) , and (-√23 , -3)
3x^2 = 69
x^2 = 23
x = ±√23
plug x^2 = 23 back into the 2nd
23 + 5y^2 = 68
5y^2 = 45
y^2 = 9
y = ±3
so 4 solutions:
(√23,3) , (√23, -3) , (-√23 , 3) , and (-√23 , -3)