Posted by Jon on Thursday, February 21, 2008 at 1:54pm.
Find the exact solution(s) of the system: x^2/4y^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

Algebra  Damon, Thursday, February 21, 2008 at 5:58pm
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(x1)
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
(z3)(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)

Algebra(from what I gathered, it's D)  Jon, Thursday, February 21, 2008 at 6:31pm
Right? PLEASE say yes

Algebra  drwls, Thursday, February 21, 2008 at 7:22pm
Yes, the answer is D
There is also an imaginary solution
x = 0, y = +/ i
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