Show that the tangent plane of the of the cone intersects the cone in a line - help in communicating idea correctly.

Question

The question I have is directly related to the following question that has been answered:

Show that the tangent plane of the cone $z^2=x^2+y^2$ at (a,b,c)$\ne$0 intersects the cone in a line

The original expression is $z^{2} = x^{2} + y^{2}$ and $(a,b,c) \neq (0,0,0)$ is the main condition.

I went about solving this question in the same exact way as the author of the original to the extent that setting the surface equal to the tangent plane I obtained

$$(x-a)^{2} + (y-b)^{2} - (z-c)^{2} = 0$$

So what I envisioned happening based on a previous exercise I had done was that I would be able to solve the above expression for a specific $(x,y,z)$. I could then put this into $f(x,y,z) = x^{2} + y^{2} - z^{2}$ and would have the equation of a line come out of this. But this won't be the case in this situation. Assuming I understand things correctly the equation of a line in $\mathbb{R}^{3}$ should be a vector of the form $t(x,y,z)$ for some scalar $t \in \mathbb{R}$. I know the mechanics is correct, but I'm missing how to express the idea of the intersecting line properly. How would I do this in this situation?

Answer 1

The equation of the (upside down) cone is: $$ f(x,y,z) = x^2+y^2-z^2 = 0 $$ Note that the top of the cone is at the origin $(0,0,0)$. Take another point $(a,b,c)\ne(0,0,0)$ at the surface of the cone: $$ a^2+b^2-c^2 = 0 $$ The equation of the OP's line in $\mathbb{R}^{3}$ is required to be of the form: $$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} t $$ So the vector $(a,b,c)$ should preferrably be normed: $$ a^2+b^2+c^2 = 1 $$ Two equations with three unknowns; we have one degree of freedom. A suitable solution is: $$ \begin{cases} a = \cos(\phi)/\sqrt{2} \\ b = \sin(\phi)/\sqrt{2} \\ c = 1/\sqrt{2} \end{cases} $$ To be interpreted as a vector to any point (angle $\phi$) on the circle $\,x^2+y^2=1/2\,$ at height $\,z=1/\sqrt{2}\,$. The end-result is: $$ \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \frac{1}{\sqrt{2}} \begin{bmatrix} \cos(\phi) \\ \sin(\phi) \\ 1 \end{bmatrix} t $$ EDIT. Before I forget, this line is, of course, in the tangent plane. $$ \vec{\nabla} f(x=a,y=b,z=c) = \begin{bmatrix} 2x \\ 2y \\ -2z \end{bmatrix} = 2\begin{bmatrix} a \\ b \\ -c \end{bmatrix} $$ So the equation of the tangent plane is (norming factors omitted): $$ ax+by-cz=0 \quad \mbox{with} \quad \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} a \\ b \\ c \end{bmatrix} t \quad \Longrightarrow \quad (a^2+b^2-c^2)t^2=0 $$ which is the same as saying that $(a,b,c)$ is at the surface of the cone.

Answer 2

As was noted in a comment directly under the linked question, $(x-a)^2 + (y-b)^2 - (z-c)^2 = 0$ is the equation of another cone. The intersection of the new cone and the original cone is the line you are looking for, and so is the intersection of the new cone and the plane, but then we're right back where we started, trying to find the intersection of two surfaces.

The key thing to keep in mind here is that the object you are looking for (which you expect to be a line) is the object described by the two equations $$f(x,y,z) = x^2 + y^2 - z^2 =0$$ and $$g(x,y,z) = 2ax+2by-2cz-a^2-b^2+c^2 = 0$$ when they are simultaneously satisfied. Note the very important "$= 0$" in each equation, without which you don't actually have the equation of either the cone or the tangent plane. The first $=$ on each line is just telling us the correspondence between two different forms of the left side of the equation -- either using the name of a function that is applied to $x,$ $y,$ and $z$ or explicitly spelling out the expression in terms of $x,$ $y,$ and $z$ that you get after you apply the function.

It is true that any point that simultaneously satisfies those two equations will also satisfy the equation $f(x,y,z) = g(x,y,z),$ but by reducing the two equations to a single equation and discarding the original equations you lose information.

So let's go back to solving the simultaneous equations \begin{align} x^2 + y^2 - z^2 &= 0, \\ 2ax+2by-2cz-a^2-b^2+c^2 &= 0. \end{align} You might note that you can simplify the second equation by using the fact that $a^2 + b^2 = c^2.$ (This was pointed out also in an answer to the linked question.)

There are various ways to solve a simultaneous equation with quadratic terms. In this case you can use the second equation to express $y$ in terms of $x$ and $z$ and plug this into the first equation, reducing it to an equation of two variables. If you treat $z$ temporarily as a given value, you have a quadratic equation in $x$ that you can solve. This will tell you what value(s) $x$ can have for any given value of $z.$ Keeping in mind always that $a^2 + b^2 = c^2,$ you should find that $x$ is proportional to $z$, and (recalling how $y$ is written in terms of $x$ and $z$, and working out the consequences) so is $y$, and therefore you have a line through the origin.