Tangents And Normals

Question

Given the curve y=18/x

Find the value of m for which the line y=mx+12 is tangent to the curve and the co-ordinates of the point of contact.

Answer 1

Eugene's answer is absolutely correct.

 

 

However, the solution may be easier to understand if you consider

there are really 2 equations in 2 unknowns:

 

equation 1: -18/a² = m   (the slopes must be equal)

 

equation 2: 18/a = ma + 12  (the intersection point must satisfy both equations)

 

If you substitute equation 1 into equation 2, you get Eugene's equation:

 

18/a = (-18/a²)(a) + 12

Answer 2

Let (a,b) be the point of contact; let m be the value such that the line y = mx + 12 is tangent to the curve y = 18/x. Since the line and curve meet at (a,b), then b = 18/a and b = ma + 12. Thus,

 

(*) 18/a = ma + 12

 

Furthermore, m is the slope of the curve y = 18/x at the point (a,b), i.e., m = dy/dx|_(x,y)=(a,b). So m = -18/x²|_(x,y)=(a,b) = -18/a². Now, substituting the expression of m into (*),

 

18/a = (-18/a²)(a) + 12

18/a = -18/a + 12

36/a = 12

a = 3

 

Since a = 3, then m = -18/3² = -2. Now b = 18/a = 6. Hence, (a,b) = (3,6).