Problem 515
Medium Tangents, Normals, and Asymptotes

Three curves are given by C_1colon y=f(x), C_2colon

Three curves are given by $C_{1}\colon y=f(x)$, $C_{2}\colon y=\varphi(x)$, $C_{3}\colon y=\dfrac{f(x)+\lambda\varphi(x)}{1+\lambda}$, where $\lambda$ is a real parameter and $f,\varphi$ are differentiable. Show that the tangents to these three curves drawn at points with the same abscissa $x=a$ meet at a single point.

Explanation of Method

A general theorem, of which Problem 514 is a special case.

Step 1: The three tangents at $x=a$

Write $f_{0}=f(a)$, $\varphi_{0}=\varphi(a)$, $f_{0}'=f'(a)$, $\varphi_{0}'=\varphi'(a)$. Then

\[ \ell_{1}\colon\ y = f_{0}'x + (f_{0}-af_{0}'), \qquad \ell_{2}\colon\ y = \varphi_{0}'x + (\varphi_{0}-a\varphi_{0}'). \tag{1,2} \]

The value and derivative of $y_{3}=\dfrac{f+\lambda\varphi}{1+\lambda}$ at $x=a$ are the corresponding weighted averages, so

\[ \ell_{3}\colon\ y = \frac{f_{0}'+\lambda\varphi_{0}'}{1+\lambda}\,x + \frac{(f_{0}-af_{0}')+\lambda(\varphi_{0}-a\varphi_{0}')}{1+\lambda}. \tag{3} \]

Step 2: The key observation: $\ell_{3}$ is a weighted average of $\ell_{1}$ and $\ell_{2}$

Comparing (1)–(3) term by term, the slope of $\ell_{3}$ is exactly $\dfrac{k_{1}+\lambda k_{2}}{1+\lambda}$ and its intercept is exactly the same weighted average of the intercepts of $\ell_{1}$ and $\ell_{2}$. Hence, as functions of $x$,

\[ \ell_{3}(x) = \frac{\ell_{1}(x)+\lambda\,\ell_{2}(x)}{1+\lambda} \qquad \text{for every } x. \tag{4} \]

Step 3: The point common to $\ell_{1}$ and $\ell_{2}$ lies on $\ell_{3}$

Let $S=(x^{*},y^{*})$ be the intersection of $\ell_{1}$ and $\ell_{2}$ (it exists and is unique whenever $f_{0}'\neq\varphi_{0}'$, i.e.\ whenever the two tangents are not parallel), so $\ell_{1}(x^{*})=\ell_{2}(x^{*})=y^{*}$. Evaluating (4) at $x^{*}$:

\[ \ell_{3}(x^{*}) = \frac{\ell_{1}(x^{*})+\lambda\,\ell_{2}(x^{*})}{1+\lambda} = \frac{y^{*}+\lambda y^{*}}{1+\lambda} = y^{*}. \tag{5} \]

So $S$ lies on $\ell_{3}$ as well. $\blacksquare$

Step 4: Verification

\[f(x)=x^3-2x,\ \varphi(x)=\sin x,\ a=0.7,\ \lambda=2.3\text{: } \ell_3(x^*)-y^*=0 \text{ (numerically)}\ \checkmark\]

\[\text{Same functions, } a=-1.1,\ \lambda=0.4\text{: } \ell_3(x^*)-y^*=0\ \checkmark\]


Result

Final answer:

\[\boxed{\,\text{Since } \ell_{3}=\frac{\ell_{1}+\lambda\ell_{2}}{1+\lambda},\text{ every point common to } \ell_{1} \text{ and } \ell_{2} \text{ also lies on } \ell_{3}\,}\]

Remark

Remark. Problem 514 is the special case $f(x)=e^{x}$, $\varphi(x)=e^{-x}$, $\lambda=1$, since $\tfrac12(e^{x}+e^{-x})=\dfrac{f(x)+1\cdot\varphi(x)}{1+1}$; the concurrency proved there is thus an instance of this general fact, and the argument above (a weighted average of two lines meets each of them only where they already meet each other) is the honest reason it works, rather than a coincidence tied to the exponential.