I am studying for my math final and our prof gave us a review but with no solutions… I am totally lost for this problem… We didn't even cover these types of problems in class so I don't know if I'll need to know how to do it but right now I have no clue. If anyone could help I would appreciate it.
Question: Show that the cone $z^2=ax^2 +by^2$ is a ruled surface.
I understand that a ruled surface is a surface composed of straight lines but that is as far as my knowledge goes for this question… Again any help is appreciated.
Best Answer
We know that by a cone we mean a geometric figure in $\mathbb R^3$ generated by a moving straight line going via a fixed curve $C$ and a fixed point not on the curve. As, you noted intuitively, the line can move freely. Let $$C:~~\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ Let $O$ be the vertex of the cone and let the curve is in the plane $z=c$ so if $P(x,y,z)\in C$, drawing a line going through $O$ and $P$ we get $$x/x_0=y/y_0=z/z_0$$ as the line equation where $Q(x_0,y_0,z_0)$ is the intersection of the line with curve. Then, the cone has the formula: $$\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=0$$ Choosing $a=c$, we get $x^2+y^2=z^2$. I hope this explanation be a good hint. Of course besides to other answers.