+0  
 
0
47
3
avatar+622 

 

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
 #1
avatar+94453 
+2

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)

 

Therefore....set 

 

(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
 #2
avatar+622 
+1

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
 #3
avatar+94453 
+1

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

 

 

cool cool cool

CPhill  Dec 31, 2018

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