Expert: Paul Klarreich - 5/11/2009

Question
Can you please help me with this problem,

Use implicit differentiation to show that the equation of the tangent to the curve y^2=kx at (x₀,y₀) is y₀*y=1/2k(x+x₀).

Answer
Questioner:   Mario Hernandez
Country:  United States
Category:  Calculus
Private:  No
 
Subject:  Implicit differentiation
Question:  Can you please help me with this problem,

Use implicit differentiation to show that the equation of the tangent to the curve y^2=kx at (x₀,y₀) is y₀*y=1/2k(x+x₀).
................................................
Hi, Mario,  

Those special symbols cause trouble.  I will write your problem as:

the equation of the tangent to the curve y^2=kx at (x0,y0) is y0 y=1/2k(x+x0)

To find the equation of the tangent line is a standard thing:

1. Differentiate to find m.
2. Use m, x0, y0 in the point-slope form:
  y - y0 = m(x - x0)
.................

If  y^2 = kx,

2y dy/dx = k

dy    k
-- = ---
dx   2y

Now at (x0,y0), m = k/(2y0)

Ready to go:

y - y0 = k/2y0(x - x0)

Multiply through by  2y0

2y0y - 2y0^2 = kx - kx0

But we have a point on the curve, don't we?  So

y0^2 = kx0

2y0y - 2kx0 = kx - kx0

2y0y = kx + kx0

and you can finish up.