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anonymous
 5 years ago
How do u solve {x^2+y^2=63
x^23y^2=27
anonymous
 5 years ago
How do u solve {x^2+y^2=63 x^23y^2=27

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there are a couple of ways to solve this, but the most efficient way is to roughly see how the graph is and find out how many intersections there are going to be

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the first eqn. is a circle with radius sqrt(63) and the second eqn. is a hyperbola

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as you can easily see , both of the conics have the center located at the origin, so the number of intersection points can vary from none to 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let's put the hyperbola in the standard form \[{x^2 \over 27}  {y^2 \over 9} =1\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0from here we can observe that this hyperbola opens sideways, and the verteces are located sqrt(27) units right and left, 3 units up and down

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oops, my bad, only sqrt(27) units right to left. the 3 units up and down tells us the asymptotes so never mind about that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since the verteces are located sqrt(27) units from the origin, and the circle has a radius of sqrt(63) we can see that the hyperbola crosses the circle four times

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0knowing this is actually important, because you don't want to waste time solving the system of eqn. only to find out there were no solutions.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0anyway, now we will start solving this algebraically

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the circle eqn can give us\[x^2 = 63y^2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if I substitute this into the hyperbola eqn \[(63  y^2)  3y^2 = 27\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0as you can see, everything in the eqn is in terms of y, so we can solve it without any trouble

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since \[4y^2 = 44 \] \[y = \pm 4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if you plug this into the circle, \[x^2 + (\pm 4)^2 = 63\] implies that \[x = \pm \sqrt{47}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0unless my calculations are wrong, the four points that the circle and the hyperbola intersect are \[(\pm \sqrt{47},\pm 4)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0wow, I did all this and skull is not even online :( oh well, let me know if anyone found this useful :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\left\{x\to 3 \sqrt{6},y\to 3\right\},\left\{x\to 3 \sqrt{6},y\to 3\right\},\left\{x\to 3 \sqrt{6},y\to 3\right\},\left\{x\to 3 \sqrt{6},y\to 3\right\} \] A plot is attached. Used the following Mathematica statement to solve it \[\text{Solve}\left[\left\{x^2+y^2==63,x^23 y^2==27\right\},\{x,y\}\right] \] Plot expression follows: Plot[%87, {x, 10, 10}, AspectRatio > Automatic] where %87 = \[\left\{\sqrt{63x^2},\sqrt{63x^2},\frac{\sqrt{27+x^2}}{\sqrt{3}},\frac{\sqrt{27+x^2}}{\sqrt{3}}\right\} \] yuki, Excellent advise with regard to having a plot to look at the solutions.
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