Find the exact solution(s) of the system: x^2/4-y^2=1 and x=y^2+1
A)(4,sqrt3),(4,-sqrt3),(-4,sqrt3),(-4,-sqrt3)
B)(4,sqrt3),(-4,sqrt3)
C)(2,1),(2,-1),(4,sqrt3),(4,-sqrt3)
D)(4,sqrt3),(4,-sqrt3)
To find the exact solutions of the given system of equations, we need to solve the two equations simultaneously. Let's start by solving the second equation for x.
Given equation: x = y^2 + 1
Now, substitute this value of x into the first equation and solve for y:
(x^2/4) - y^2 = 1
((y^2 + 1)^2)/4 - y^2 = 1
(y^4 + 2y^2 + 1)/4 - y^2 = 1
Now, multiply every term by 4 to get rid of the fraction:
y^4 + 2y^2 + 1 - 4y^2 = 4
Combine like terms:
y^4 - 2y^2 - 3 = 0
Now, let's factorize this quadratic equation:
(y^2 - 3)(y^2 + 1) = 0
Therefore, either (y^2 - 3) = 0 or (y^2 + 1) = 0
Solving the first equation:
y^2 - 3 = 0
y^2 = 3
Taking the square root of both sides:
y = sqrt(3) or y = -sqrt(3)
Now, substitute these values back into the second equation to find the corresponding values of x:
For y = sqrt(3):
x = (sqrt(3))^2 + 1
x = 3 + 1
x = 4
Therefore, one solution is (x, y) = (4, sqrt(3)).
For y = -sqrt(3):
x = (-sqrt(3))^2 + 1
x = 3 + 1
x = 4
Therefore, another solution is (x, y) = (4, -sqrt(3)).
So, the correct answer is option C) (2, 1), (2, -1), (4, sqrt(3)), (4, -sqrt(3)).