Find the exact solution(s) of the system: x^2/4-y^2=1 and x=y^2+1

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)).