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)
I don't know
I have to sketch such a problem to see what I am doing.
x^2 /2^2 -y^2/1^2 = 1
is a hyperbola centered on the origin and opening right and left.
the right vertex is at x = 2 and the left at -2
the slope of the arms at infinity is +/- 2
The other curve is a parabola
y^2 = x - 1
y = +/- sqrt(x-1)
vertex at x = 1, y = 0
opens right
So I should get two solutions (if there are any solutions) at the same x and y the same absolute value, opposite signs.
Now
(y^2+1)^2/4 - y^2 = 1
let z = y^2
(z+1)^2 - 4 z = 4
z^2 + 2 z + 1 - 4 z = 4
z^2 -2 z -3 = 0
(z-3)(z+1) = 0
z = 3 or z = -1
but z = y^2
y^2 = 3
y = +/- 3 THOSE ARE THE ONES
y^2 = -1
y = +/- sqrt(-1) imaginary (corresponds to the left half of the hyperbola where the parabola never goes)
Right? PLEASE say yes
Yes, the answer is D
There is also an imaginary solution
x = 0, y = +/- i
To find the exact solution(s) of the given system of equations:
1. Start by substituting the value of x from the second equation into the first equation:
x = y^2 + 1
Substitute this into the first equation:
(y^2 + 1)^2/4 - y^2 = 1
2. Simplify the equation by expanding:
(y^4 + 2y^2 + 1)/4 - y^2 = 1
(y^4 + 2y^2 + 1 - 4y^2)/4 = 1
(y^4 - 2y^2 + 1)/4 = 1
3. Multiply both sides of the equation by 4 to eliminate the fraction:
y^4 - 2y^2 + 1 = 4
4. Rearrange the equation:
y^4 - 2y^2 - 3 = 0
5. Factor the quadratic equation:
(y^2 - 3)(y^2 + 1) = 0
6. Set each factor equal to zero and solve for y:
y^2 - 3 = 0 --> y^2 = 3 --> y = ±sqrt(3)
y^2 + 1 = 0 --> y^2 = -1 --> No real solution for y
7. Substitute the values of y 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
For y = -sqrt(3):
x = (-sqrt(3))^2 + 1 --> x = 3 + 1 --> x = 4
Therefore, the solution to the system of equations is (4, sqrt(3)) and (4, -sqrt(3)), which corresponds to option B) (4, sqrt(3)), (-4, sqrt(3)).