Warning: this is an htmlized version!
The original is across this link,
and the conversion rules are here.
% (find-angg "LATEX/2016-2-GA-P2.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2016-2-GA-P2.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2016-2-GA-P2.pdf"))
% (defun e () (interactive) (find-LATEX "2016-2-GA-P2.tex"))
% (defun u () (interactive) (find-latex-upload-links "2016-2-GA-P2"))
% (find-xpdfpage "~/LATEX/2016-2-GA-P2.pdf")
% (find-sh0 "cp -v  ~/LATEX/2016-2-GA-P2.pdf /tmp/")
% (find-sh0 "cp -v  ~/LATEX/2016-2-GA-P2.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2016-2-GA-P2.pdf
%               file:///tmp/2016-2-GA-P2.pdf
%           file:///tmp/pen/2016-2-GA-P2.pdf
% http://angg.twu.net/LATEX/2016-2-GA-P2.pdf
\documentclass[oneside]{book}
\usepackage[colorlinks]{hyperref} % (find-es "tex" "hyperref")
%\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{pict2e}
\usepackage{color}                % (find-LATEX "edrx15.sty" "colors")
\usepackage{colorweb}             % (find-es "tex" "colorweb")
%\usepackage{tikz}
%
\usepackage{edrx15}               % (find-angg "LATEX/edrx15.sty")
\input edrxaccents.tex            % (find-angg "LATEX/edrxaccents.tex")
\input edrxchars.tex              % (find-LATEX "edrxchars.tex")
\input edrxheadfoot.tex           % (find-dn4ex "edrxheadfoot.tex")
\input edrxgac2.tex               % (find-LATEX "edrxgac2.tex")
%
\begin{document}

\catcode`\^^J=10
\directlua{dednat6dir = "dednat6/"}
\directlua{dofile(dednat6dir.."dednat6.lua")}
\directlua{texfile(tex.jobname)}
\directlua{verbose()}
%\directlua{output(preamble1)}
\def\expr#1{\directlua{output(tostring(#1))}}
\def\eval#1{\directlua{#1}}
\def\pu{\directlua{pu()}}

\directlua{dofile "edrxtikz.lua"} % (find-LATEX "edrxtikz.lua")
\directlua{dofile "edrxpict.lua"} % (find-LATEX "edrxpict.lua")
%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end
\pu

\def\V(#1){\VEC{#1}}




%   ____      _                    _ _           
%  / ___|__ _| |__   ___  ___ __ _| | |__   ___  
% | |   / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \ 
% | |__| (_| | |_) |  __/ (_| (_| | | | | | (_) |
%  \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/ 
%                                                

{\setlength{\parindent}{0em}
\footnotesize
\par Geometria AnalÃtica
\par PURO-UFF - 2016.2
\par P2 - 18/jan/2017 - Eduardo Ochs
\par Respostas sem justificativas nÃo serÃo aceitas.
\par Proibido usar quaisquer aparelhos eletrÃ∧nicos.
% \par VersÃo: 14/mar/2016
% \par Links importantes:
% \par \url{http://angg.twu.net/2015.2-C2.html} (pÃgina do curso)
% \par \url{http://angg.twu.net/2015.2-C2/2015.2-C2.pdf} (quadros)
% \par \url{http://angg.twu.net/LATEX/2015-2-C2-material.pdf}
% \par {\tt eduardoochs@gmail.com} (meu e-mail)
}

\bsk
\bsk

\setlength{\parindent}{0em}
\def\T(Total: #1 pts){{\bf(Total: #1 pts)}}
\def\T(Total: #1 pts){{\bf(Total: #1)}}
\def\B       (#1 pts){{\bf(#1 pts)}}
% Usage:
% 1) \T(Total: 2.34 pts) Foo
% a) \B(0.45 pts) Bar

Lembre que uma equaÃÃo de cÃ∧nica à uma equaÃÃo da forma $ax^2 + bxy +
cy^2 + dy + ey + f = 0$; $4+(x+y)(x-y)=5y$ nÃo à uma equaÃÃo de cÃ∧nica
mas à equivalente a uma: $x^2-y^2-5y+4=0$.

Nas questÃes 3 e 4 vamos usar a abreviaÃÃo $[\text{equaÃÃo}] =
\setofxyzst{\text{equaÃÃo}}$.

\bsk

1) \T(Total: 4.0 pts) FaÃa esboÃos das cÃ∧nicas com as equaÃÃes abaixo.
Algumas delas sÃo degeneradas.
%
$$
\begin{tabular}[t]{rl}
  a) (0.1 pts) & $xy=0$ \\
  b) (0.1 pts) & $(x+y)(x-y)=0$ \\
  c) (0.1 pts) & $(y-1)(y-2)=0$ \\
  d) (0.1 pts) & $x+y=0$ \\
  e) (0.1 pts) & $x+y=1$ \\
  f) (0.1 pts) & $(x+y)^2=1$ \\
  g) (0.1 pts) & $y+2=0$ \\
  h) (0.1 pts) & $y+2=1$ \\
\end{tabular}
\quad
\begin{tabular}[t]{rl}
  i) (0.4 pts) & $(x+y)^2+(y+2)^2=0$ \\
  j) (1.0 pts) & $(x+y)^2+(y+2)^2=1$ \\
  k) (0.4 pts) & $(x+y)(y+2)=0$ \\
  l) (1.0 pts) & $(x+y)(y+2)=1$ \\
  m) (0.4 pts) & $(\frac{x-4}{2})^2 + (\frac{y-6}{3})^2=1$ \\
\end{tabular}
$$

\bsk

2) \T(Total: 1.0 pts) Encontre os focos das elipses cujos pontos Ãbvios sÃo:

a) \B(0.5 pts) $(\pm 3, 0)$, $(0, \pm 5)$

b) \B(0.5 pts) $(4 \pm 3, 2)$, $(4, 2 \pm 5)$

\bsk
\bsk

3) \T(Total: 1.5 pts) Sejam $Ï_1=[y=3-2x]$, $Ï_2=[x+y+z=4]$, $r=Ï_1âÏ_2$.

a) \B(0.5 pts) Encontre o ponto de $r$ que tem $x=0$.

b) \B(0.5 pts) Encontre o ponto de $r$ que tem $x=1$.

c) \B(0.5 pts) Se $r'=\setofst{(x,ax+b,cx+d)}{xâ\R}$ e $r=r'$, quem
sÃo $a$, $b$, $c$ e $d$?

\bsk
\bsk

4) \T(Total: 3.5 pts) Sejam $A=(1,0,0)$, $B=(0,1,0)$, $\uu=\vec{AB}$,
$C=(1,1,0)$, $D=(1,0,1)$, $\vv=\vec{CD}$, $r=\setofst{A+t\uu}{tâ\R}$,
$r'=\setofst{C+t'\vv}{t'â\R}$.

a) \B(1.5 pts) Demonstre (algebricamente!) que $r$ e $r'$ sÃo
reversas, ou, equivalentemente, que $A$, $B$, $C$ e $D$ nÃo sÃo
coplanares.

b) \B(2.0 pts) Calcule a distÃncia entre $r$ e $r'$.


\newpage

%   ____       _                _ _        
%  / ___| __ _| |__   __ _ _ __(_) |_ ___  
% | |  _ / _` | '_ \ / _` | '__| | __/ _ \ 
% | |_| | (_| | |_) | (_| | |  | | || (_) |
%  \____|\__,_|_.__/ \__,_|_|  |_|\__\___/ 
%                                          

Gabarito:

\msk

% (find-LATEX "2016-2-GA-algebra.tex" "elipses")

\unitlength=5pt
\def\closeddot{\circle*{0.6}}

\def\pictpoint#1{\put(#1){\closeddot}}
\def\pictline#1{{\linethickness{1.0pt}\expr{Line.new(#1):pict()}}}
\def\pictlinethin#1{{\linethickness{0.2pt}\expr{Line.new(#1):pict()}}}
\def\pictLine(#1)(#2)#3{%
  \vcenter{\hbox{%
   \beginpicture(#1)(#2)%
   \pictaxes%
   \pictline{#3}
   \end{picture}%
  }}%
 }

\def\pictellipse#1{{\linethickness{1.0pt}\expr{Ellipse.new(#1):pict()}}}
\def\pictEllipse(#1)(#2)#3{%
  \vcenter{\hbox{%
   \beginpicture(#1)(#2)%
   \pictaxes%
   \pictellipse{#3}
   \end{picture}%
  }}%
 }
\def\pictEllipseF(#1)(#2)#3(#4)(#5){%
  \vcenter{\hbox{%
   \beginpicture(#1)(#2)%
   \pictaxes%
   \pictellipse{#3}
   \put(#4){\closeddot}
   \put(#5){\closeddot}
   \end{picture}%
  }}%
 }

\def\picthyperbole#1#2{{\linethickness{1.0pt}\expr{Hyperbole.new(#1):pict(#2)}}}



1a)
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictline{v(0,0), v(1,0), -2, 2}
   \pictline{v(0,0), v(0,1), -2, 2}
   \end{picture}%
 }}$
\quad
1b)
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictline{v(0,0), v(1,1), -2, 2}
   \pictline{v(0,0), v(1,-1), -2, 2}
   \end{picture}%
 }}$
\quad
1c)
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictline{v(0,1), v(1,0), -2, 2}
   \pictline{v(0,2), v(1,0), -2, 2}
   \end{picture}%
 }}$
\quad
1d)
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictline{v(0,0), v(-1,1), -2, 2}
   \end{picture}%
 }}$
\quad
1e)
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictline{v(1,0), v(-1,1), -1, 2}
   \end{picture}%
 }}$
\quad
1f)
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictline{v(1,0),  v(-1,1), -1, 2}
   \pictline{v(-1,0), v(-1,1), -2, 1}
   \end{picture}%
 }}$

\msk

1g)
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictline{v(0,-2),  v(1,0), -2, 2}
   \end{picture}%
 }}$
\quad
1h)
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictline{v(0,-1),  v(1,0), -2, 2}
   \end{picture}%
 }}$

% \msk

\def\mygrid{
   \pictlinethin{v(0,-1), v(1,0), -1, 5}  % y-2 = 1
   \pictlinethin{v(0,-2), v(1,0), -1, 5}  % y-2 = 0
   \pictlinethin{v(0,-3), v(1,0), -1, 5}  % y-2 = -1
   \pictlinethin{v(0,-1), v(1,-1), -2, 3}  % x+y = -1
   \pictlinethin{v(0, 0), v(1,-1), -2, 4}  % x+y = 0
   \pictlinethin{v(1, 0), v(1,-1), -2, 4}  % x+y = 1
}

1i)
$\vcenter{\hbox{%
   \beginpicture(-2,-4)(5,2)%
   \pictaxes%
   \mygrid
   \put(2,-2){\closeddot}
   \end{picture}%
 }}$
\quad
1j)
$\vcenter{\hbox{%
   \beginpicture(-2,-4)(5,2)%
   \pictaxes%
   \mygrid
   \pictellipse{v(2,-2), v(1,0), v(-1,1)}
   \end{picture}%
 }}$
\quad
1k)
$\vcenter{\hbox{%
   \beginpicture(-2,-4)(5,2)%
   \pictaxes%
   \mygrid
   \pictline{v(0,-2), v(1,0), -1, 5}  % y-2 = 0
   \pictline{v(0, 0), v(1,-1), -2, 4}  % x+y = 0
   \end{picture}%
 }}$
\quad
1l)
$\vcenter{\hbox{%
   \beginpicture(-2,-4)(5,2)%
   \pictaxes%
   \mygrid
   \picthyperbole{v(2,-2), v(1,0), v(-1,1), 1}{10, -4, -1/2, 1/4, 4}
   \end{picture}%
 }}$
\quad
1m)
$\vcenter{\hbox{%
   \beginpicture(-1,-1)(7,10)%
   \pictaxes%
   \pictellipse{v(4,6), v(2,0), v(0,3)}
   \end{picture}%
 }}$



\bsk

\unitlength=5pt
\def\closeddot{\circle*{0.6}}

2a) $\pictEllipseF(-4,-6)(4,6){v(0,0), v(3,0), v(0,5)}(0,4)(0,-4)$
\quad
2b) $\pictEllipseF(-1,-4)(8,8){v(4,2), v(3,0), v(0,5)}(4,6)(4,-2)$

\bsk

3a) $x=0 \;â\; y=3 \;â\; z=1 \;â\; P=(0,3,1)$

3b) $x=1 \;â\; y=1 \;â\; z=2 \;â\; P=(1,1,2)$

3c) $r'=\setofst{(x,-2x+3,1x+1)}{xâ\R}$, $a=-2$, $b=3$, $c=1$ e $d=1$

\bsk

4a) $[\Vec{AB}, \Vec{AC}, \Vec{AD}]
     = [\V(-1,1,0), \V(0,1,0), \V(0,0,1)]
     = \vsm{-1&1&0 \\ 0&1&0 \\ 0&0&1}
     = -1 \neq 0$,

portanto $A$, $B$, $C$ e $D$ nÃo sÃo coplanares.

4b) Seja $\ww=\Vec{AC}=\V(0,1,0)$. EntÃo
%
$$\begin{array}{rcl}
  d(r,r') &=& [\Vec{AB}, \Vec{CD}, \Vec{AC}] / ||\Vec{AB} Ã \Vec{CD}|| \\
          &=& [\V(-1,1,0), \V(0,-1,1), \V(0,1,0)] / ||\V(-1,1,0) Ã \V(0,-1,1)|| \\
          &=& \vsm{-1&1&0 \\ 0&-1&1 \\ 0&1&0} / ||\V(1,1,1)|| \\
          &=& 1/\sqrt{3} \\
          &=& \sqrt{3}/3 \\
  \end{array}
$$
%

\end{document}




% Local Variables:
% coding: utf-8-unix
% ee-anchor-format: "«%s»"
% End: