PROBLEM 51 (Page 113): How Many Tangent Lines to the Curve Pass Through the Point

x PROBLEM 51 (Page 113): How many tangent lines to the curve y  pass x  1 through the point (1, 2) ? At which points do these tangent lines touch the curve?

x First, check to see if the point (1, 2) is on the curve y  : x  1

1 x y(1)   2 . Thus, the point (1, 2) is not on the curve y  . 2 x  1

In general, we have the following picture:

NOTE: The point ( x, f ( x ) ) is the tangent point.

Since the two points of ( x, f ( x ) ) and ( a , b ) are two points on the tangent line, we can find the slope of the tangent line using algebra:

f ( x )  b m  tan x  a

Of course, we can find the slope of the tangent line to the graph of y  f ( x ) at the point ( x, f ( x ) ) using calculus: m tan  f ( x ) f ( x )  b x Thus, we have that f ( x )  . For this problem, f ( x )  . x  a x  1

1(x  1)  x (1) x  1  x 1 Thus, f ( x )  = = and ( x  1) 2 ( x  1) 2 ( x  1) 2

x x f ( x )  b  2  2 x  2( x  1) = x  1 = x  1 x  1 = = x  a  ( x  1) ( x  1) x  1 x  1 x  1

x  2 x  2  x  2  ( x  2) = = ( x  1)( x  1) ( x  1)( x  1) ( x  1)( x  1)

f ( x )  b 1  ( x  2) Thus, f ( x )     x  a ( x  1) 2 ( x  1)( x  1)

( x  1) ( x  1)   ( x  2)( x  1) 2 

( x  1)( x  1)  ( x  2)( x  1) 2  0 

( x  1)[ x  1  ( x  2)( x  1)]  0 

( x  1)( x  1  x 2  3x  2)  0  ( x  1)( x 2  4x  1)  0  x  1  0 or x 2  4x  1  0

x  1  0  x   1. Since the domain of the function f is all real numbers except  1 , then we do not have a tangent point at  1 .

 b  b 2  4 a c  4  16  4(1)(1) x 2  4x  1  0  x    2 a 2  4  16  4  4  12  4  2 3 = = =  2  3 2 2 2 x These are the x-coordinates of the tangent points to the graph of y  of the x  1 tangent lines that pass through the point (1, 2) . Thus, there are two tangent lines that pass through the point (1, 2) .  2  3  2  3 If x   2  3 , then f (  2  3 )  = =  2  3  1 3  1

 2  3 3  1  2 3  2  3  3 1  3  = = . Thus, one 3  1 3  1 3  1 2  1  3    2  3 ,  tangent point is   .  2   2  3  2  3 If x   2  3 , then f (  2  3 )  = =  2  3  1  1  3

2  3 2  3 1  3 2  2 3  3  3  1  3 =  = = = 1  3 1  3 1  3 1  3  2

1  3  1  3    2  3 ,  . Thus, the other tangent point is   . 2  2 

Maple commands to solve this problem and draw the graph.