Write the equations of the two asymptotes?

Question

Graph (y+3)2/9 - (x-2)2/16 = 1

Part I: Identify the coordinates of the center of this hyperbola.

center = (2,-3)

Part II: Use the values of a or b to locate the coordinates of the vertices. 
=(2,-3,-3) , ( 2,-3+3) 
=(2,-6) , (2,0)

Part III: What are the coordinates of the foci? 
=(2,-3 + or - sqrt 16+9 ) 
= ( 2, -3 + or - 5) 
= (2,-8) , (2,2) 
Part IV: Write the equations of the two asymptotes.

I need help with part 4 please!!

Jim · Answer

The way you wrote it was incorrect for asymptotes. You wrote  (y+3)(2/9) - (x-2)(2/16) = 1
Should have been:
(y+3)^2 /9 - (x-2)^2 /16 = 1 
or
(y+3)² /9 - (x-2)² /16 = 1

? · Answer

the asymptotes pass thru the center --- part 1  --  and have slopes m = +/- (b/a)  --- a = 3, b = 4
  so slopes are +/- (4/3)
   use point slope formula for EACH slope to write the 2 equations
  y - y1  =  m(x - x1)  <<<   (x1, y1)  is the center

I think you can take it from here ... Good luck.

Write the equations of the two asymptotes?

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