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Please solve this for me..

Guest Dec 31, 2014

Best Answer 

 #1
avatar+88899 
+10

Here's (b)

x^2 + xy - y^2  = 1    using implicit differentiation, we have

2x + y + xy' - 2yy'  = 0

xy' - 2yy'  = -2x - y       multiply through by - 1 on both sides

2yy' - xy'  = 2x + y

y'(2y - x) = 2x + y

y' =  [2x + y ] /  [2y - x ]      and the slope of the tangent line at (2.3) is given by

y' = [2(2) + 3] / [2(3) - 2 ]   =   [7] / [ 6 - 2]  =  7 / 4  

Here's a graph of the tangent line to the "rotated" hyperbola at the point (2,3)....https://www.desmos.com/calculator/yuxxmj7ud9

Note, that the slope of the tangent line at this point = (7/4).....just as we expected...!!!

 

CPhill  Dec 31, 2014
 #1
avatar+88899 
+10
Best Answer

Here's (b)

x^2 + xy - y^2  = 1    using implicit differentiation, we have

2x + y + xy' - 2yy'  = 0

xy' - 2yy'  = -2x - y       multiply through by - 1 on both sides

2yy' - xy'  = 2x + y

y'(2y - x) = 2x + y

y' =  [2x + y ] /  [2y - x ]      and the slope of the tangent line at (2.3) is given by

y' = [2(2) + 3] / [2(3) - 2 ]   =   [7] / [ 6 - 2]  =  7 / 4  

Here's a graph of the tangent line to the "rotated" hyperbola at the point (2,3)....https://www.desmos.com/calculator/yuxxmj7ud9

Note, that the slope of the tangent line at this point = (7/4).....just as we expected...!!!

 

CPhill  Dec 31, 2014

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