Finding tangent lines to Devil's Curve

Question

This is a fairly interesting question and I look forward to being able to solve it.

The curve's equation is given by $ y^4-96y^2 = x^4-100x^2$. I have differentiated this implicitly to get $y^1= \frac{(x^3-50x)}{(y^3-48y)}$

Now the questions asks that I find all the tangent lines at where $x = 10$. I drew the graph out and noticed that there should be several tangent lines existing at this point. Therefore, the next step I took was to find different $y$ values that satisfied the equation. I ended up with:

$ y= 0, y= \sqrt 96, y=-\sqrt 96 $

I tried subbing these points back into both the original equation and derivative to try and find the slope, but I don't believe I ended up with the right answer. I got, for the point $(10, \sqrt 96)$, a tangent equation of $y = 1.063x - 10.63$, which when plotted with the curve itself, did not look like a proper tangent.

Any help would be much appreciated! Please excuse my poor formatting.

Answer 1

Everything is fine, up to the very last little bit. The equation of a line with gradient $m$ & going through $(a,b)$ is \begin{eqnarray*} y-b=m(x-a). \end{eqnarray*} You missed $\color{red}{b}$ out. ... you should get $y=1.063x-0.8335$ (to 4 sig. fig.) or more precisely \begin{eqnarray*} y=\frac{125}{48\sqrt{6}}x-\frac{98}{48\sqrt{6}} \end{eqnarray*}

... https://www.desmos.com/calculator/ktdrcadu3z

Answer 2

The curve $$f(x,y) = y^4-96y^2 - x^4 + 100x^2 = 0$$ at $x = 10$ has precisely three tangent lines, one corresponding to each real-valued solution of $f(10,y) = 0$; namely, $$0 = y^4 - 96y^2 = y^2 (y^2-96) = y^2(y - 4 \sqrt{6})(y + 4 \sqrt{6}).$$ This corresponds to the three points $$(x,y) \in \{ (10,0), (10, 4\sqrt{6}), (10,-4\sqrt{6}) \}.$$ Since $f(x,-y) = f(x,y)$ for all $(x,y)$, this suggests that the tangent line at $(10,0)$ is vertical, and the other two points of tangency are related in that one is the negative of the other. Indeed, $$\frac{dy}{dx} = \frac{x(x^2 - 50)}{y(y^2 - 48)},$$ and when $y = 0$ and $x \not\in \{0, \pm 5 \sqrt{2}\}$, the slope is undefined (i.e., vertical). For the other two points, it is easy to compute $$\left[\frac{dy}{dx}\right]_{(x,y) = (10,\pm 4\sqrt{6})} = \frac{500}{\pm 192\sqrt{6}} = \pm \frac{125}{48\sqrt{6}}.$$ Thus the vertical tangent line has equation $$x = 10,$$ and the other two tangent lines are $$y \mp 4 \sqrt{6} = \pm \frac{125}{48\sqrt{6}}(x - 10) .$$

It is instructive to determine the set of all points satisfying $f(x,y) = 0$ for which the tangent line to $f$ is vertical; i.e. the complete set $(x_0, y_0)$ satisfying $dx/dy = 0$. How many are there?