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How do you solve the conic of 16x^2 - y^2 +16y -128 =0

Guest May 4, 2015

Best Answer 

 #1
avatar+87309 
+15

16x^2 - y^2 +16y -128 =0   add 128 to both sides

16x^2 - y^2 +16y  = 128     complete the square on y

16x^2 - (y^2 - 16y + 64) = 128 - 64    simplify

16x^2 - (y - 8)^2  = 64       divide both sides by 64

x^2 / 4  - (y - 8)^2  / 64   = 1

 

This is a hyperbola with a center of (0, 8)......a = 2 and b = 8...the focal points are at (±√68, 8)

Here's a graph ........https://www.desmos.com/calculator/hagrdi3shl

 

 

  

CPhill  May 4, 2015
 #1
avatar+87309 
+15
Best Answer

16x^2 - y^2 +16y -128 =0   add 128 to both sides

16x^2 - y^2 +16y  = 128     complete the square on y

16x^2 - (y^2 - 16y + 64) = 128 - 64    simplify

16x^2 - (y - 8)^2  = 64       divide both sides by 64

x^2 / 4  - (y - 8)^2  / 64   = 1

 

This is a hyperbola with a center of (0, 8)......a = 2 and b = 8...the focal points are at (±√68, 8)

Here's a graph ........https://www.desmos.com/calculator/hagrdi3shl

 

 

  

CPhill  May 4, 2015
 #2
avatar+92805 
+10

Thanks Chris, these are really interesting questions.  There is much I could learn from your answer  :)

Melody  May 5, 2015
 #3
avatar+87309 
+10

Thanks, Melody.........these conics have many facets......I barely touched the surface.....LOL!!!!

 

  

CPhill  May 5, 2015

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