Tuesday, September 22, 2009

Double tangent line

Find the equation of the line that is tangent to the curve
y = x4 − 14x3 + 69x2 at two points.


  1. The answer is y = 140x - 100

    Let y = ax + b be the equation of the double tangent line. Then x^4 − 14x^3 + 69x^2 − ax − b has two double roots, so it can be factored as (x-p)^2 (x-q)^2.

    Expanding and equating coefficients gives the following system of equations.

    2p + 2q = 14

    p^2 + 4pq + q^2 = 69

    2pq^2 + 2p^2q = −a

    p^2 q^2 = −b

    Solving the first pair of equations yields (p,q) = (2,5) or (5,2). Substituting into the second pair of equations yields a = 140 and b = −100.

  2. Found this a slightly different way. Since the slope m, of the line y=mx+b, is tangent to f(x)=x^4 − 14x^3 + 69x^2, m must be the same as the first derivative of f(x) at these intersections. Since f(x)=f'(x)x-b, we can infer that b = f(x)-f'(x)*x.

    I then used a graphing calculator to graph the [m,b] values of the tangent line at each point along the graph: x(t) = [ f'(x), f(x)-f'(x)*x ]. This parametric line intersects itself at [140,-100], which gives the line y=140x-100.