Find the equation of the line that is tangent to the curve
y = x4 − 14x3 + 69x2 at two points.
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The answer is y = 140x - 100
ReplyDeleteLet 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.
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.
ReplyDeleteI 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.