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!

  • $\begingroup$ Dear mary, the part of your sentence "... an artist sitting at 0 an element..." makes no sense: is English your mother tongue? Also, why do you ask this question and where does it come from? $\endgroup$ Sep 2, 2012 at 9:46
  • $\begingroup$ I am needing to better understand Algebraic Geometry and this is a question from a textbook $\endgroup$
    – mary
    Sep 2, 2012 at 22:46
  • $\begingroup$ @GeorgesElencwajg-I can't imagine that the native language of the OP is relevant over and above whether the question is comprehensible. I've now edited it so that I think it is so. $\endgroup$ Sep 2, 2012 at 23:28
  • 1
    $\begingroup$ Dear @Kevin, of course the native language of the OP is relevant: it often happens that you cannot express yourself correctly in a foreign language and a mistake in syntax might make your question incomprehensible. By the way, English is not my native language and, in case this was not perfectly clear, my remark was meant to be friendly and sympathetic toward a fellow non-anglophone. $\endgroup$ Sep 4, 2012 at 7:25
  • 1
    $\begingroup$ Completely unrelated: does somebody know why the software allows a user who has 56 points of reputation to offer a bounty of +100 ? I thought bounties were subtracted from the reputation. $\endgroup$ Sep 4, 2012 at 7:43

2 Answers 2


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.

  • $\begingroup$ Can you please provide an explicit formula? That'll help me out greatly. Thanks! $\endgroup$
    – mary
    Sep 10, 2012 at 0:50
  • $\begingroup$ I guess since you have the other answer you don't still need it? $\endgroup$ Sep 10, 2012 at 10:39

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]:


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:


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


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:


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 $B$:


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 $c\in range [-1,10]$:


Circle/Parabola, $c = 1$:


Circle/Ellipse, $c < 1$:


Circle/Hyperbola, $c > 1$:


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

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



  • $\begingroup$ Heh, Ben beat me to it, but I already had all this nice animations :P, so I figured I'll add it. $\endgroup$
    – Realz Slaw
    Sep 10, 2012 at 3:48
  • $\begingroup$ you did a terrific job! I would also award you the bounty if I can $\endgroup$
    – mary
    Sep 10, 2012 at 8:45
  • $\begingroup$ @mary: the lesson is, don't award bounties early :P $\endgroup$ Sep 10, 2012 at 10:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .