If a variable point $P$ moves such that the line passing through $P$ and $Q(0,0,2)$ makes an angle $60^{\circ}$ with the $z$-axis then find the locus of $P$. Also find the locus of intersection of locus of $P$ with the plane $x+y+z=1$.
The first part of the problem is pretty easy. The locus I got is $$x^2+y^2-3(z-2)^2=0.$$
Now for the second part, I substituted $z=1-x-y$ in the above equation and got that the locus is a hyperbola.
But the thing is that the answer given in the solution is ellipse. I asked by teacher about this and he did this by some angle method. He said for the locus to be an ellipse the angle at which it should cut has to be greater than $60^{\circ}$, and for a hyperbola it has to be less than $60^{\circ}$. When he did this problem by his method, be also got the locus as a hyperbola and not an ellipse.
Why is an ellipse the correct locus and not a hyperbola?
Best Answer
I would say that hyperbola is the right answer: the angle reasoning gives a good image about what's happening; the equation you found gives a cone with vertex in Q and the linear equation defines a plane passing trough $(0,0,1)$ parallel to the line $\{x=-y,\;z=0\}$ and making a $\frac{\pi}{4}$ angle with the $z$-axis; this produces an hyperbola because the plane meets the cone on two "sides".
Probably your professor just got confused with the $<,>$ signs since later the answer given was the same as yours.
You can visualize this using Desmos or Geogebra3D or, if you want to draw it yourself, using only a "slice" of the objects so that you can draw it on a plane: for example you could set $x=0$ and you would get $y^2-3{(z-2)}^2=0$ for the cone, that becomes just two lines, and $y+z=1$ for the plane, that becomes just a line; the line (ex-plane) intersect the cone in two points on two sides and this behaviour is typical of hyperbolas.
Substituting as you said you get $x^2+y^2-3{(1+x+y)}^2=0$ and it is not simple to see right away what this is. If in that equation you use the substitutions $u=x+y,\;v=x-y$ (that also imply $x=\frac{u+v}{2},\;y=\frac{u-v}{2}$) you get the equation $u^2+v^2-6{(1+u)}^2=0$ that makes it simpler to see that it's an hyperbola.