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In $\mathbb{R}^3$ say we have the 2 planes $A=\{z=1\}$ and $B=\{x=1\}$. A line through 0 meeting $A$ at $(x,y,1)$ meets $B$ at $(1,y/x,1/x).$ Consider the map $\phi: A \rightarrow B$ defined by $(x,y) \mapsto (y' = y/x, z' = 1/x)$.

I'm trying to figure out the image under $\phi$ of

1) the line $ax = y + b$; the pencil of parallel lines $ax = y + b$ with fixed $a$ and variable $b$;

2) circles $(x-1)^2 + y^2 = c$ for variable $c,$ distinguishing the 3 cases $c>1, c = 1,$ and $c< 1$.

and to imagine the above as a perspective drawing by an artist sitting at $(0,0,0)$ and drawing figures from the plane $A$ on the plane $B$.

What happens to the points of the 2 planes where $\phi$ and $\phi^{-1}$ are undefined? Thanks!

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    Dear @Ben: of cour$s$e! I $s$hould have thought o$f$ this...Anyway, thanks a lot for your explanation.2012-09-09

2 Answers 2

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To answer your last question first, notice that a line through $0$ meeting $A$ at $(0,y,1)$ does not meet $B$ at all. This explains why $\phi$ is undefined in such cases. Correspondingly, pick any point on $B$ with $z = 0$ and any line through the origin and that point is wholly within the $xz$-plane, so will never hit $x = 1$, so is not the projection of any point on $A$, so $\phi^{-1}$ is undefined.

To understand how lines on $A$ work, think of lines as the intersection of planes. More specifically, for each line $\lambda$ in $A$ there is a unique plane $C$ through the origin such that $\lambda$ is the intersection of $A$ with $C$. Then the image under $\phi$ must be the intersection of $C$ with $B$ (since any "projection ray" from the origin through $\lambda$ lies in the plane $C$). Now, this intersection will be a line in $B$ (assuming the line was not $\{x = 0, z = 1\}$, in which case there is no intersection). So lines project to lines. Once we have that fact, it's easy to compute which line it is: just project any two points of $\lambda$, and join them up. If you really need an explicit formula, just ask.

Circles are a little trickier. Substitute $x=1/z\prime$ and $y=y\prime/z\prime$ into the equation, and get: \[\frac{1}{z^2}(y^2 + (1-z)^2)=c\]. What does this actually mean? Well, let's rearrange a little: \[\begin{align} \frac{1}{z^2}(y^2 + 1 - 2z + z^2) &= c \\ y^2 + 1 - 2z + z^2 &= cz^2 \\ y^2 - 2z + (1-c)z^2 &= -1 \end{align}\]. At this point I want to divide by $1-c$ to complete the square, so I'm going to have to distinguish the $c=1$ case. In that case, we get \[\frac{1}{2}(y^2 + 1)=z\], which is a parabola. Otherwise: \[\begin{align} y^2 + (1-c)(z^2 - \textstyle{\frac{2}{1-c}}z) &= -1 \\ y^2 + (1-c)((z-\textstyle{\frac{1}{1-c}})^2 - \textstyle{\frac{1}{(1-c)^2}}) &= -1 \\ y^2 + (1-c)(z-\textstyle{\frac{1}{1-c}})^2 &= \textstyle{\frac{1}{1-c}} - 1 \\ y^2 + (1-c)(z-\textstyle{\frac{1}{1-c}})^2 &= \textstyle{\frac{c}{1-c}} \end{align}\]. For $c < 1$, this is an ellipse, while for $c > 1$, it is a hyperbola.

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    I guess since $y$ou have the other answer you don't still need it?2012-09-10
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1) the line $ax = y + b$; the pencil of parallel lines $ax = y + b$ with fixed $a$ and variable $b$;

Lets transform this to B in the following way:

$\begin{eqnarray*} && ax &=& y + b && \text{Given}\\ \times && 1/x &=& 1/x && \text{Divide both sides by } x\\ && a &=& \frac y x + \frac b x && \text{Simplify}\\ && y' &=& \frac y x && \text{Given under }\phi\\ && a &=& y' + \frac b x && \text{Replacing }\frac y x \text{ with } y'\\ && z' &=& \frac 1 x && \text{Given under }\phi\\ && a &=& y' + bz' && \text{Replacing }\frac b x = b\cdot \frac 1 x \text{ with } b\cdot z'\\ && y' &=& [-b]z'+[a] && \text{Simplifying to slope-intercept form}\\ \end{eqnarray*}$

Next, lets rewrite $ax=y+b$ in the [slope-intercept form][hyperlink at bottom]:

$y=[a]x+[-b]$

In this form, [a] is the slope and [-b] can be interpreted as:

the y-intercept of the line, the y-coordinate where the line intersects the y-axis

― [wikipedia/Slope#Algebra][hyperlink at bottom]

Now we can see, while $a$ is the slope of the line in $A$, in $B$, $[a]$ is the how far off the x-axis the line intercepts. Thus we begin to see a relationship between the two lines; as the line rotates in $A$, it will rise and fall in $B$. To be more precise: as the slope becomes greater in $A$ (line rotates counter clockwise), the line in will rise in $B$.

A similar relationship exists for the rise and fall of the line in $A$ to the slope of the line in $B$. Since the rise/fall of the line in $A$ is determined by $[-b]$, when the line rises, its image in $B$ will rotate. So in direct answer: with variable $b$, as $b$ becomes greater, the line will fall in $A$, and the slope will become steeper in $B$, clockwise, and the line will approach being vertical, falling on the right side of the y-axis. When $b$ becomes lesser, the line will rise in $A$ and rotate counter-clockwise in $B$, and as $-b$ increases, the line will approach being vertical. So a pencil of parallel lines will look like a slice through a fan, where the lines in B all rotate around a fixed point on the y-axis, all meeting through $(a,0)$. The undefined values for this equation will be a vertical line in $A$ and $B$, which results in an infinite slope. To quote wikipedia:

The [slope-intercept form] fails for a vertical line, parallel to the y axis (see Division by zero), where the slope can be taken as infinite, so the slope of a vertical line is considered undefined.

― wikipedia/Slope#Algebra[hyperlink at bottom]

To approach this case, the line in $A$ will have to rise or fall to infinity ($\pm\infty$), and the slope in $B$ will approach vertical. Similarly, when the slope of the line in $A$ approaches vertical, the line in $B$ will rise or fall to $\pm\infty$.

Some images: Scroll to bottom for links.

2) circles $(x-1)^2 + y^2 = c$ for variable $c,$ distinguishing the 3 cases $c>1, c = 1,$ and $c< 1$.

$(x-1)^2 + y^2 = c$

$\frac {(x^2 -2x +1)} {x^2} + y^2 = c\\$

$\begin{eqnarray*} (x-1)^2 + y^2 &=& c && \text {Given}\\ \times \frac 1 {x^2} && &&\text{Divide both sides by } x\\ \frac {(x^2 -2x +1)+y^2} {x^2} &=& c \cdot \frac 1 {x^2} && \text{Simplification}\\ 1 - \frac {2}{x} + \frac 1{x^2} + \frac {y^2} {x^2} &=& c \cdot \frac 1 {x^2} && \text{Simplification}\\ 1 - 2\cdot\frac {1}{x} + \left[\frac 1{x}\right]^2 + \left[\frac {y} {x}\right]^2 &=& c \cdot \left[\frac 1 {x} \right ]^2 && \text{Separate terms that look like our transformation}\\ 1 - 2\cdot z' + \left[z'\right]^2 + \left[y'\right]^2 &=& c \cdot \left[z'\right ]^2 && \text{Plug in } y' \text{ and } z' \text{ into their respective places}\\ \end{eqnarray*}$

Solving for ${y'}^2$: $ {y'}^2 = (c-1){z'}^2 + 2z' - 1 $

Now we want to get this into an equation form that is descriptive of its graph.

When $c=1$, we get ${y'}^2 = 0{z'}^2 + 2z' - 1$, implying ${y'}^2 = 2z' - 1$, and thus $z' = y'^2 - 1$, which is the form of an east opening parobola.

For $c \neq 1$:

Lets "[complete the square][hyperlink at bottom]": $\begin{eqnarray*} {y'}^2 &=& (c-1){z'}^2 + 2z' - 1\\ {y'}^2 &=& (c-1)\left[{z'}^2 + \left[\frac {2} {c-1}\right]2z'\right] - 1\\ 0 &=& + \left[\frac {1}{\left(c-1\right)}\right] - \left[\frac {1}{\left(c-1\right)}\right]\\ 0 &=& + \left[\frac {\left(c-1\right)}{\left(c-1\right)^2}\right] - \left[\frac {1}{\left(c-1\right)}\right]\\ {y'}^2 &=& (c-1)\left[{z'}^2 + \left[\frac {2} {c-1}\right]2z'\right] - 1 + \left[\frac {\left(c-1\right)}{\left(c-1\right)^2}\right] - \left[\frac {1}{\left(c-1\right)}\right] \\ {y'}^2 &=& (c-1)\left[{z'}^2 + \left[\frac {2} {c-1}\right]2z' \left[\frac {1}{\left(c-1\right)^2}\right]\right] - 1 - \left[\frac {1}{\left(c-1\right)}\right] \\ {y'}^2 &=& (c-1)\left[{z'}^2 + \left[\frac {2} {c-1}\right]2z' \left[\frac {1}{\left(c-1\right)}\right]^2\right] - 1 - \left[\frac {1}{\left(c-1\right)}\right] \\ {y'}^2 &=& (c-1)\left[z'+\left[\frac {1}{\left(c-1\right)}\right]\right]^2 - 1 - \left[\frac {1}{\left(c-1\right)}\right] &&\text{Completing the square}\\ {y'}^2 &=& (c-1)\left[z'+\left[\frac {1}{\left(c-1\right)}\right]\right]^2 - \frac {\left(c-1\right)} {\left(c-1\right)} - \left[\frac {1}{\left(c-1\right)}\right] \\ {y'}^2 &=& (c-1)\left[z'+\left[\frac {1}{\left(c-1\right)}\right]\right]^2 - \frac {c} {c-1}\\ 1 &=& \frac {\left[z'+\left[\frac {1}{c-1)}\right]\right]^2} { \frac {c} {(c-1)^2}} - \frac {{y'}^2} { \frac {c} {c-1}} \end{eqnarray*}$

$ \frac {\left[z'+\left[\frac {1}{c-1)}\right]\right]^2} { \frac {c} {(c-1)^2}} - \frac {{y'}^2} { \frac {c} {c-1}} = 1 $ Now, this is in the form of the [ellipse][hyperlink at bottom] and [hyperbola][hyperlink at bottom] in their equations.

Ellipse equation, from [wikipedia/Ellipse#Canonical_form][hyperlink at bottom]: ($(X_c,Y_c)$ represents the center): $ \frac {\left(x - X_c\right)^2} a + \frac {\left(y - Y_c\right)^2} b = 1 $ And an "East-West opening hyperbola" hyperbola, centered at $(h,k)$, from [wikipedia/Hyperbola#Cartesian_coordinates][hyperlink at bottom]:

$ \frac {\left(x - h\right)^2} a - \frac {\left(y - k\right)^2} b = 1 $

For an ellipse, the $+ \left[\frac 1 {c-1} \right]$ section in the $z'$ part is translation, while the denominators relate to the shape.

So when is this an ellipse and when is it a hyperbola? When $\frac c {c-1}$ is negative, than this becomes of the form of an ellipse (with a $+$ in between the nomials). If $\frac c {c-1}$ is positive (when $c > 1$) then the 2nd nomial subtracts from the first, which is of the hyperbola's form (east-west opening hyperbola).

Images at bottom.

The circle and ellipse both become undefined when $c < 0$.

For the circle, $c < 0$ would mean a negative radius, which results in imaginary (square root of a negative number) $x,y$.

For the ellipse, $c < 0$ would result in the form:$-\frac x a - \frac y b = 1$, which is really $\frac x a + \frac y b = -1$, which would also result in imaginary $x,y$.

In the following images, the xy plane and yz plane are super-imposed over each-other. Pink is the image on $A$, and green is the transformed image on $B$.

Animation of line, as b varies:

http://i.stack.imgur.com/JXf5d.gif 

Line on $A$ raised high, slope on $B$ approaching vertical counter-clockwise:

http://i.stack.imgur.com/Z4Jem.png 

If the green slope would be vertical, the pink line would be at (intercept the y-axis at) $\pm\infty$.

Line on $A$ low, slope on $B$ approaching vertical clockwise:

http://i.stack.imgur.com/9dEjT.png 

If the green slope would be vertical, the pink line would be at (intercept the y-axis at) $\pm\infty$.

Pencil of lines on $A$, corresponding fan of lines on $B$:

http://i.stack.imgur.com/4fqKN.png 

As the pink line moves among its parallel pencil positions, the lines on $B$ rotate around $(a,0)$

Animation of circle/projected circle for varying $c\in range [-1,10]$:

http://i.stack.imgur.com/0R4ij.gif 

Circle/Parabola, $c = 1$:

http://i.stack.imgur.com/xws0J.png 

Circle/Ellipse, $c < 1$:

http://i.stack.imgur.com/c3Khi.png 

Circle/Hyperbola, $c > 1$:

http://i.stack.imgur.com/roA1f.png 

Geogebra file for general/projected line/projected point/inverse projected point:

http://www.geogebratube.org/material/show/id/16973 You can use this to see where things are undefined by dragging around point P on plane A and seeing the corresponding P' on plane B. 

Geogebra file for projected circle

http://www.geogebratube.org/material/show/id/16972 

Links:

http://en.wikipedia.org/wiki/Linear_equation#Slope.E2.80.93intercept_form http://en.wikipedia.org/wiki/Slope#Algebra http://en.wikipedia.org/wiki/Completing_the_square http://en.wikipedia.org/wiki/Ellipse#Canonical_form http://en.wikipedia.org/wiki/Hyperbola#Cartesian_coordinates 
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    @mary: the lesson is, don't award bounties early :P2012-09-10