#1**+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**+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