How prove this equation $A^2+B^2=C^2+D^2$

Question

define:

plane $W:Ax+By+Cz+D=0$ and the hyperboloid of one sheet $U:x^2+y^2-z^2=1$ if $$W\bigcap U=l_{1},W\bigcap U=l_{2}$$

where $l_{1},l_{2}$ be two straight lines

show that :$$A^2+B^2=C^2+D^2$$

My try: since $$\begin{cases} Ax+By+Cz+D=0\\ x^2+y^2-z^2=1 \end{cases}$$ then $$(Ax+By+D)^2=C^2(x^2+y^2-1)$$ then $$(A^2-C^2)x^2+(B^2-C^2)y^2+2ABxy+2BDy+2ADx+D^2+C^2=0$$

Follow is user44197 idea $$(A^2-C^2)x^2+(2ABy+2AD)x+(B^2-C^2)y^2+D^2+C^2=0$$ $$\Delta (y)=(2ABy+2AD)^2-4(A^2-C^2)[(B^2-C^2)y^2+D^2+C^2]$$ $$\Delta=4A^2B^2y^2+8A^2BDy+4A^2D^2-4(A^2B^2-A^2C^2-B^2C^2+C^4)y^2-4(A^2D^2+A^2C^2-C^2D^2-C^4)$$ so $$\Delta(y)=4[C^2(A^2+B^2-C^2)y^2+2A^2BDy+C^2(C^2+D^2-A^2)]$$ then I can't.Thank you very much!

The mistake is in the term term $((B^2-C^2)y^2+D^2+C^2)$. It should be $((B^2-C^2)y^2+2BDy+D^2+C^2)$ — user44197, Dec 26 '13 at 04:54

score 1 · Accepted Answer · answered Dec 26 '13 at 03:54

1

You need to make use of the fact that the intersection is a pair of straight lines.

Look at the last equation as a quadratic in $x$, i.e of the form $a x^2+b x +c=0$ Its discriminant is $b^2-4ac$ is given by $$ \Delta(y) = 4\,{C}^{2}\,\left( {D}^{2}+2\,y\,B\,D-{y}^{2}\,{C}^{2}+{C}^{2}+{y}^{2}\,{B}^{2}+{y}^{2}\,{A}^{2}-{A}^{2}\right) $$ which is a quadratic in $y$. Now taking its discriminant (as a quadratic in $y$) gives $$ \Delta = 64\,{C}^{4}\,\left( C-A\right) \,\left( C+A\right) \,\left( {D}^{2}+{C}^{2}-{B}^{2}-{A}^{2}\right) $$ This has to be zero to get the intersection to be straight lines. Ruling $C=A$ and $C=-A$ and $C=0$ you get the desired answer.

answered Dec 26 '13 at 03:54

user44197

9,730

Hello,when $C=A$,why have $A^2+B^2=C^2+B^2=C^2=D^2\Longrightarrow B=D$? – math110 Dec 26 '13 at 04:00
C^2=A^2 is degenerate case. In this case, we have no $x^2$ term. Similarly if we start by eliminating $y$ first we will get the degenerate case $B^2=D^2$. Both these are degenerate and so is $C=0$, and $D=0$. The key is you have to make use of the fact that the intersection is two straight lines – user44197 Dec 26 '13 at 04:10
Hello,I don't understand you mean? can you post full solution? – math110 Dec 26 '13 at 04:21
you mean $\Delta(x)\Longrightarrow \Delta=0\Longrightarrow B=D?$,I think this is not ture – math110 Dec 26 '13 at 04:22
No! $\Delta=0$ gives 4 possibilities: $C=0$, $C=A$, $C=-A$, and $A^2+B^2=C^2+D^2$. Only the last makes sense. The other 3 are degenerate (meaning the intersection is no longer a quadratic in $x$ and $y$. – user44197 Dec 26 '13 at 04:32
Sorry, I see the typo in my first reply. It should be $B^2=C^2$. In any case, you can't have $C=0$, $C = \pm A$. So you are left with $A^2+B^2=C^2+D^2$. – user44197 Dec 26 '13 at 04:34
Hello,I think $$\Delta(y)=4[C^2(A^2+B^2-C^2)y^2+2A^2BDy+C^2(C^2+D^2-A^2)]$$ – math110 Dec 26 '13 at 04:36
let us continue this discussion in chat – user44197 Dec 26 '13 at 04:39
Yes,I have edit,plese see my calculations ,where is my wrong – math110 Dec 26 '13 at 04:47
See my edit of your original post with the correction. – user44197 Dec 26 '13 at 04:56

score 1 · Answer 2 · edited Dec 26 '13 at 10:47

Here's an answer with a bit of a geometric flavor.

Clearly, the hyperboloid contains no lines that are parallel to any of the coordinate axes; it also contains no points $(x,y,z)$ such that $x^2 + y^2 < 1$. Consequently, an embedded line $\ell$ must pass through a point $P$ on the $xy$-plane, which must in fact lie on the unit circle; we can write $P = (\cos\theta, \sin\theta, 0)$ for some $\theta$. Moreover, the projection of $\ell$ into the $xy$-plane must be tangent to the unit circle, so its direction vector, $v$, has the form $(\sin\theta,-\cos\theta,c)$ for some $c$. That is, the line has this parameterization: $$\ell : (\cos\theta, \sin\theta, 0 ) + t (\sin\theta,-\cos\theta, c)$$ and we have $$0 = x^2 + y^2 - z^2 - 1 = (\cos\theta+t\sin\theta)^2+(\sin\theta-t\cos\theta)^2+(ct)^2-1 =t^2(c^2-1)$$ which must hold for all $t$, so that $c=\pm 1$.

Since $P$ is on the plane: $$A \cos\theta + B \sin\theta + D = 0 \qquad \implies \qquad D = -\left(A \cos\theta+ B \sin\theta\right)$$ Since $v \perp (A,B,C)$: $$v\cdot(A,B,C) = A\sin\theta - B\cos\theta \pm C = 0 \qquad \implies \qquad C = \mp \left( A \sin\theta - B \cos\theta \right)$$

Therefore, $$C^2 + D^2 = A^2 + B^2 \qquad (\star)$$

Note: We have shown that $(\star)$ holds whenever the plane meets the hyperboloid in even a single line. This is consistent with the fact that $(\star)$ removes only a single degree of freedom in the plane equation.

How prove this equation $A^2+B^2=C^2+D^2$

2 Answers2