I am trying to solve the following problem and explaining it to my students. However, my students (as well as I) think of my two assumptions contradictory.

> Given 2 distinct curves $C_1: y=f(x)=e^{6x}$ and $C_2: y=g(x)=ax^2$ where $a>0$. The objective is to find the range of $a$ such that there exists 2 tangents, each is tangent to both given curves.

I have solved it as follows.

  1. Let $(t,e^{6t})$ be "a single point" on $C_1$ through which a tangent passes.

    The slope of the tangent is $6e^{6t}$ that is obtained from the first derivative of $f(x)=e^{6x}$ at $x=t$.

    The tangent is $$ y-e^{6t} = 6e^{6t}(x-t) $$

  2. Let $(u,au^2)$ be a point on $C_2$ through which the tangent passes. It means that the tangent and $C_2$ have only one intersection point $(u,au^2)$.

    We have, $$ au^2 -e^{6t} = 6e^{6t}(u-t) $$ By arranging it, we have a quadratic equation in $u$. $$ au^2 -6e^{6t}u + e^{6t}(6t-1) =0 $$

    In order to make the quadratic equation have twin roots, its discriminant must be equal to zero as follows. $$ (-6e^{6t})^2-4\times a\times e^{6t}(6t-1) =0 $$

  3. There will be 2 tangents if there are 2 roots $t_1$ and $t_2$ for the equation in $t$ below. $$ (-6e^{6t})^2-4\times a\times e^{6t}(6t-1) =0 $$

    We can interpret finding the roots of the equation as finding the abscissa $t$ of the intersection of $\alpha(t)=a$ and $\beta(t)=\frac{9e^{6t}}{6t-1}$.

    Using derivative, we can determine the minimum value of $\beta$ which is $9e^2$. Thus $a>9e^2$ is the required range.

My confusion

In the first step I must assume there is a single $t$ but in the last (the 3rd step) step I have to assume there are 2 roots $t_1$ and $t_2$. It looks inconsistent, doesn't it? How can we explain it better without causing such a confusion?

  • $\begingroup$ Where have you used the condition that the line is tangent to $C_2$? $\endgroup$ Commented Apr 26, 2015 at 11:26
  • $\begingroup$ I thought that was for finding two solutions, not for proving tangent to $C_2$. $\endgroup$ Commented Apr 26, 2015 at 13:31
  • $\begingroup$ Oh, in the 2nd step you assume only one intersection point with the parabola. That's equivalent to the line being tangent to $C_2$. $\endgroup$ Commented Apr 27, 2015 at 7:02
  • $\begingroup$ Why must you "assume there is a single $t$" as opposed to just "a $t$"? $\;$ $\endgroup$
    – user5109
    Commented Apr 27, 2015 at 7:34
  • $\begingroup$ @RickyDemer I think what is confusing for the OP is that the line has to intersect $C_1$ in a single point, while in the last step there has to be at least two choices for that single point. $\endgroup$ Commented Apr 27, 2015 at 8:31

2 Answers 2


Here's how I solved the problem:

The tangent line to $y=e^{6x}$ at $x=x_e$ is $y=6e^{6x_e}(x-x_e)+e^{6x_e}$.

The tangent line to $y=ax^2$ at $x=x_q$ is $y=2ax_q(x-x_q)+ax_q^2$.

The two lines are equal if and only if the slope and intercepts are equal. This gives the system of equations: $$6e^{6x_e}=2ax_q$$ $$(1-6x_e)e^{6x_e} = -ax_q^2.$$ Substitute $e^{6x_e}=\frac{1}{3}ax_q$ from the first into the second to get $$\frac{1}{3}(-1+6x_e)=x_q.$$ Then substituting this back into the first equality above gives $$a = \frac{9e^{6x_e}}{6x_e-1}.$$ From here it's as you have above: Calculus provides the minimum is $a=9e^2$.

When $a>9e^2$, there are (at least) two solutions for $x_e$ since the graph positively diverges at both $x_e=1/6$ and as $x_e \to \infty$.

  • $\begingroup$ Your solution is correct but you did not answer my question. Thanks for participating. $\endgroup$ Commented May 2, 2015 at 5:26
  • $\begingroup$ I thought the question was "How can we explain it better?" This method certainly avoids the potentially inconsistent assumptions. $\endgroup$
    – Aeryk
    Commented May 4, 2015 at 14:26

I "feel" there are no common tangent (in red,with a small and large; in green, the exponential).

enter image description here

  • 3
    $\begingroup$ When $a>9e^2$ there are two common tangents. When $a=9e^2$ there is one common tangent. (Also for some negative values of $a$ there is one common tangent.)) Your large $a$ in the picture is not large enough. Try $a=70$. $\endgroup$ Commented Apr 27, 2015 at 20:30
  • $\begingroup$ You are right I guess. Remember I said only "I feel". My examples were large and small just to illustrate with graphics (its were the large too small and the small too large because if not you get two perpendicular red lines and nothing to illustrate the reason of "feeling"). Anyway try to be really convinced of your assertion. Sorry for bad english. $\endgroup$
    – Piquito
    Commented Apr 28, 2015 at 14:40

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