Algebra

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

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asked by Jon
  1. 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)

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    posted by Damon
  2. Right? PLEASE say yes

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    posted by Jon
  3. Yes, the answer is D
    There is also an imaginary solution
    x = 0, y = +/- i

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    posted by drwls

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