# 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.

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Instead of trying to show that the cone is composed of lines you only need show that if point $p=(x_0,y_0,z_0)$ is on the cone, you can find a line through $p$ such that the whole line is on the cone. – John Douma Apr 27 '13 at 5:26
Have you tried making a nice illustrating picture? – Babak S. Apr 27 '13 at 5:33

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.

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Me, too, Babak + 1 – amWhy Apr 28 '13 at 0:40
I appreciate the detailed answer but I am still confused. We didn't cover this topic in the course and I just wrote the practice final and again another question about this topic was on it. This time the question was: The hyperboloid of one sheet $x^2+y^2-z^2 = 1$ is called a ruled surface, which is to say that it is made up of straight lines. Prove this by showing that for any $\theta$, the line $$\frac{x - cos\theta}{sin\theta} = \frac{y-sin\theta}{-cos\theta} = z/1$$ lies entirely on the surface... I would really appreciate more help if you don't mind. – user68203 Apr 29 '13 at 6:37
@user68203: I am at your service for the next hours, if you don't mind – Babak S. Apr 29 '13 at 8:07
Never mind I figured it out. Thanks for the help. – user68203 Apr 29 '13 at 16:05

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.

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thank you for the help – user68203 Apr 29 '13 at 16:27

$z^2=ax^2+by^2$

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!

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+1 you were the first inspiring one. – Babak S. Apr 27 '13 at 6:21