show that a cone is a ruled surface 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.
 A: 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.
A: $z^2=ax^2+by^2$
Write
$x = \frac z{\sqrt a}\cos\theta$
$y=\frac z{\sqrt b}\sin\theta$
Fix an $(x,y,z)$ on the cone.
You retrieve $\theta$ using above and you get the straight line $(t,\frac t{\sqrt a}\cos\theta,\frac t{\sqrt b}\sin\theta)$ ($\theta$ fixed) through $(x,y,z)$ lying on the cone.
Since$(x,y,z)$ was arbitrary, we're done!
A: This is easy to see if we switch to cylindrical coordinates. Substitute $x=r\cos \theta$ and $y=r\sin \theta$ to get $$z^2=ar^2\cos^2\theta + br^2\sin^2\theta = r^2(a\cos^2\theta + b\sin^2\theta)$$
This gives $$z=\pm r(a\cos^2\theta + b\sin^2\theta)^{\frac{1}{2}}$$
For a cone, $a$ and $b$ must be positive so for any $\theta$ this gives a pair of lines through the origin in the $r$-$z$ plane. If $p=(x_0,y_0,z_0)$ is not the origin, then $z_0=r(a\cos^2\theta_0 + b\sin^2\theta_0)^{\frac{1}{2}}$ for some $\theta_0$ which is on the line $$z=r(a\cos^2\theta_0 + b\sin^2\theta_0)^{\frac{1}{2}}$$
Therefore, this cone is a ruled surface.
