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After graphing the C1 and playing around on desmos i figured out a= 3

but can someone explain to me why please because i dont see the correlation between the new asymptotes and (x-3)^2

thank you 

 Dec 30, 2018

This is a hyperbola centered at the origin  intersecting the x axis


The equations of the asymptotes are


y = (2/3)x     and    y = (-2/3)x


Notice that  these lines intersect at the hyperbola's center.....if we set these equal, we can verify that the origin is the center


(2/3)x = (-2/3)x

x = -x

2x = 0

x = 0


And putting this value of x  into either linear equation gives y =0


So...the center of C1  is (0,0)




(2/3)x - 2  = (-2/3)x + 2       


(4/3)x = 4     Implies that   x = 3


And putting this value back into either equation  gives y  = (2/3)(3) - 2  = 0


So....the center of C2 is (3, 0).....so    C2 is the translation of C1 -  3 units to the right   


So....as you found YEEEEEET.....a =  3.....!!!!


cool cool cool

 Dec 31, 2018
edited by CPhill  Dec 31, 2018

ohhh i was finding it weird as the equation for asymptotes are +-b/a but as i followed up to the part where u explained how asymptotes can find origin i immediately clicked on. thanks a lot!!!

YEEEEEET  Dec 31, 2018

OK....this one was kinda' interesting....!!!



cool cool cool

CPhill  Dec 31, 2018

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