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September 16, 2014

September 16, 2014

Posted by **Jon** on Thursday, February 21, 2008 at 1:54pm.

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 5:58pmI 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)

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

- Algebra -
**drwls**, Thursday, February 21, 2008 at 7:22pmYes, the answer is D

There is also an imaginary solution

x = 0, y = +/- i

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