# algebra 2

1. Find the exact solutions of x^2 - (y-12)^2 = 144 and y = -x^2

2. Solve the system: y^2 - 4x^2 = 4 and y = 2x
Solving this myself I got no solution, but someone who helped me with this question got the answer x= + i sqrt 2, -i sqrt 2

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1. 1. They do not intersect
http://www.wolframalpha.com/input/?i=plot+x%5E2+-+(y-12)%5E2+%3D+144+and+y+%3D+-x%5E2

2. Just like in #1, there is no REAL solution but both have an imaginary or complex solution
y^2 - 4x^2 = 4 and y = 2x , use substitution
(2x)^2 - 4x^2 = 4
0 = 4 , contradiction, no solution at all

interesting situation,
y^2/4 - x^2 = 1
or
x^2 - y^2/4 = -1 is a vertical hyperbola
the asymptotes are x-y/2 = 0 and x + y/2 = 0
x - y/2 = 0 ----> y = 2x <------- the given line, hyperbolas reach their asymptotes at infinity

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2. 1. x^2 - (y-12)^2 = 144, and Y = -x^2.
x^2 - (y^2 - 24y + 144) = 144,
Replace x^2 with -y:
-y - y^2 + 24y -144 = 144,
-y^2 + 23y - 144 = 144,
-y^2 + 23y - 288 = 0,
Y = (-23 +- sqrt(529 - 1152))/(-2) ,
Y = (-23 +- 24.96i)/(-2) = 11.5 - 12.5i, and 11.5 + 12.5i.
So we have 2 imaginary solutions.

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