음함수의 미분법

The slope of the tangent line is, of course, given by the derivative, so let's start by finding $\frac{\mathrm{d}y}{\mathrm{d}x}$ of both shapes.

$$\begin{aligned} \\ \text{For the circle, we differentiate implicitly} \\ 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{y} \\ \text{For the ellipse, we also differentiate implicitly:} \\ 2x+6y\frac{\mathrm{d}y}{\mathrm{d}x}&=0 \\ \text{and solve for $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x}$} \\ \frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{3y} \\ \text{What we want is a value of $x$ where both derivatives are equal. However, they might have different values of $y$, so let's let $y_1$ be the $y$-values associated with $x$ on the circle, and let $y_2$ be the $y$-values associated with $x$ on the ellipse. That is, $x^2+ y_1^2=1$ and $x^2+3y_2^2=1$. For the slopes at $(x,y_1)$ on the circle and $(x,y_2)$ on the ellipse to be equal, we need:} \\ -\frac{x}{y_1}&=-\frac{x}{3y_2}\\ x\left(\frac{1}{y_1}-\frac{1}{3y_2}\right)&=0 \\ \text{So $x=0$ or $y_1=3y_2$. Let's think about which $x$-values will have a $y$-coordinate of the circle be three times as large as a $y$-coordinate of the ellipse. If $y_1=3y_2$, $(x,y_1)$ is on the circle, and $(x,y_2)$ is on the ellipse, then $x^2+y_1^2 =x^2+(3y_2)^2=1$ and $x^2+3y_2^2=1$. In this case:} \\ x^2+9y_2^2&=x^2+3y_2^2\\ 9y_2^2&=3y_2^2\\ y_2&=0\\ x&=\pm 1 \end{aligned}$$

We need to be a tiny bit careful here: when $y=0$, $y'$ is not defined for either curve. For both curves, when $y=0$, the tangent lines are vertical (and so have no real-valued slope!). Two vertical lines are indeed parallel.

So, for $x=0$ and for $x=\pm1$, the two curves have parallel tangent lines.