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
So I should get two solutions (if there are any solutions) at the same x and y the same absolute value, opposite signs.
(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
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