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Given two functions $f(x)=e^{6x}$ and $g(x)=ax^2$ where $a>0$. The objective is to find a condition for $a$ such that there are exactly 2 lines, each is tangent to both of the given functions.

My attempt

Let $L$ be one of the tangent lines. Also let $(\alpha,f(\alpha))$ and $(\beta,g(\beta))$ be the tangent points of $L$ to $f$ and $g$, respectively.

Based on the constraints given above, we have the following simultaneous equations.

\begin{align} g(\beta)-f(\alpha) &= f'(\alpha)(\beta-\alpha)\\ f(\alpha)-g(\beta) &= g'(\beta)(\alpha-\beta) \end{align}

With some manipulations, I obtain

$$ a=\frac{9e^{6\alpha}}{6\alpha-1} $$

And with calculus, the minimal positive value of $a$ is $9e^2$. Thus the final answer is $a>9e^2$.

I wonder how this way leads me to the conclusion there are 2 tangent lines?

Edit: I also have a relation $6\alpha = 3\beta +1$ but it seems "useless".

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1 Answers1

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Your $\alpha$ satisfies the equation

$$ 9e^{6\alpha} = a(6\alpha - 1) $$

Let

$$ h(x) = 9e^{6x} + a(1-6x) $$

Since $$\lim_{x\to \infty} h(x) = \infty$$

In order for $h(x)$ to have two roots, we require $$ \lim_{x\to -\infty} h(x) = \infty $$

which is true for any $a > 0$, and $h(x)$ has to have a negative minimum value

First, we find $$ h'(x) = 6 (9e^{6x} - a) $$

which is zero at some $x_0$ where $$ 6x_0 = \ln\left( \frac{a}{9} \right) $$

Plugging this back in to our function, we get $$ h(x_0) = a\left[2 - \ln\left(\frac{a}{9}\right) \right] $$

Since $a > 0$, this gives $$ 2 - \ln\left(\frac{a}{9}\right) < 0 $$

or $a > 9e^2$

If you didn't have $a > 0$, it's possible to only have one solution, as $h(x) \to \infty$ for $a < 0$.

For $0 \le a < 9e^2$, there is no solution, and for $a = 9e^2$ there is exactly one.

Dylan
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