% (find-angg "LATEX/2017-2-GA-P2.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2017-2-GA-P2.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2017-2-GA-P2.pdf"))
% (defun b () (interactive) (find-zsh "bibtex 2017-2-GA-P2; makeindex 2017-2-GA-P2"))
% (defun e () (interactive) (find-LATEX "2017-2-GA-P2.tex"))
% (defun u () (interactive) (find-latex-upload-links "2017-2-GA-P2"))
% (find-xpdfpage "~/LATEX/2017-2-GA-P2.pdf")
% (find-sh0 "cp -v  ~/LATEX/2017-2-GA-P2.pdf /tmp/")
% (find-sh0 "cp -v  ~/LATEX/2017-2-GA-P2.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2017-2-GA-P2.pdf
%               file:///tmp/2017-2-GA-P2.pdf
%           file:///tmp/pen/2017-2-GA-P2.pdf
% http://angg.twu.net/LATEX/2017-2-GA-P2.pdf

% «.gab-1»	(to "gab-1")
% «.gab-2a-2j»	(to "gab-2a-2j")
% «.gab-2k-2s»	(to "gab-2k-2s")
% «.gab-3»	(to "gab-3")
% «.gab-4»	(to "gab-4")

\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}
%
% (find-dn6 "preamble6.lua" "preamble0")
%\usepackage{proof}   % For derivation trees ("%:" lines)
%\input diagxy        % For 2D diagrams ("%D" lines)
%\xyoption{curve}     % For the ".curve=" feature in 2D diagrams
%
\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}}
\def\und#1#2{\underbrace{#1}_{#2}}



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

{\setlength{\parindent}{0em}
\footnotesize
\par Geometria Analítica
\par PURO-UFF - 2017.2
\par P2 - 11/dez/2017 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Diagramas muito ambíguos {\sl serão} interpretados errado.
\par Proibido usar quaisquer aparelhos eletrônicos.

}

\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 equação de cônica é uma equação da forma $ax^2 + bxy +
cy^2 + dx + ey + f = 0$; $4+(x+y)(x-y)=5y$ não é uma equação de cônica
mas é equivalente a uma: $x^2-y^2-5y+4=0$. E o truque pra gente se
livrar das duas raízes quadradas em $√A + √B = C$ ou $√A - √B = C$ é:
%
$$\begin{array}{rcl}
  √A + √B = C &⇒& C^2(C^2 - 2(A+B)) + (A-B)^2 = 0 \\
  √A - √B = C &⇒& C^2(C^2 - 2(A+B)) + (A-B)^2 = 0 \\
  \end{array}
$$

% Nas questões 3 e 4 vamos usar a abreviação $[\text{equação}] =
% \setofxyzst{\text{equação}}$.


\bsk

1) \T(Total: 1.5 pts) Use as fórmulas acima para transformar
%
$$d((x,y),(-2,0))-d((x,y),(2,0))=2$$
%
numa equação de cônica e verifique se os seguintes pontos obedecem a
equação original e a equação da cônica: $(1,0)$, $(2,3)$, $(-2,3)$,
$(-1,0)$.

\bsk

% \def\myu {\frac{x-4}{2}}
% \def\myv {     y-\frac{x}{2}  }
\def\myu {y-x/2-3}
\def\myv {y+x/2-3}
\def\myuu{\und{\textstyle\myu}{u}}
\def\myvv{\und{\textstyle\myv}{v}}

2) \T(Total: 4.5 pts) Faça esboços das cônicas com as equações abaixo.
Algumas delas são degeneradas. Em todos os itens abaixo considere que
$u=2y$ e $v=x+y$ --- ou que $u$ e $v$ são {\sl abreviações} para $2y$
e $x+y$.
%
$$
\begin{tabular}[t]{rl}
  a) (0.1 pts) & $x^2+y^2=1$ \\
  b) (0.1 pts) & $x^2+y^2=4$ \\
  c) (0.1 pts) & $x^2+y^2=0$ \\
  d) (0.1 pts) & $xy=1$ \\
  e) (0.1 pts) & $xy=4$ \\
  f) (0.1 pts) & $xy=0$ \\
  g) (0.1 pts) & $xy=-1$ \\
  h) (0.1 pts) & $x^2=y$ \\
  i) (0.1 pts) & $x=y^2$ \\
  j) (0.1 pts) & $x^2=y^2$ \\
\end{tabular}
\quad
\begin{tabular}[t]{rl}
  k) (0.5 pts) & $u(u-1)=0$  \\
  l) (0.5 pts) & $v(v-1)=0$  \\[4pt]
  m) (0.2 pts) & $u^2+v^2=0$ \\
  n) (0.3 pts) & $u^2+v^2=1$ \\[4pt]
  o) (0.2 pts) & $uv=0$      \\
  p) (0.4 pts) & $uv=1$      \\
  q) (0.4 pts) & $uv=-1$     \\[4pt]
  r) (0.5 pts) & $u^2=v$     \\
  s) (0.5 pts) & $u=v^2$     \\
\end{tabular}
$$

\bsk

3) \T(Total: 1.5 pts) Dê uma parametrização para a reta que pertence
aos planos $x+y+z=3$ e $x+2z=3$.

\bsk

4) \T(Total: 2.5 pts) Sejam:
%
$$\begin{array}{rcl}
  r &=& \setofst{(t,t,0)}{t∈\R} \\
  s &=& \setofst{(2t,-2t,4)}{t∈\R} \\
  r' &=& \setofst{(t,t,0)}{t∈\R} \\
  s' &=& \setofst{(t,0,3-t}{t∈\R} \\
  \end{array}
$$

a) \B(0.1 pts) Calcule $d(r,s)$ no olhômetro.

a) \B(0.9 pts) Calcule $d(r,s)$ pela fórmula.

c) \B(1.5 pts) Calcule $d(r',s')$ pela fórmula.


\newpage



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

{\bf Gabarito} (versão não revisada)

\msk

%  _ 
% / |
% | |
% | |
% |_|
%    
% «gab-1» (to ".gab-1")

1) $d((x,y),(-2,0))-d((x,y),(2,0))=2$

$⇒ \sqrt{(x+2)^2+y^2} - \sqrt{(x-2)^2+y^2} = 2$. Sejam $C=2$,

$A = (x+2)^2+y^2 = x^2+4x+4+y^2$,

$B = (x-2)^2+y^2 = x^2-4x+4+y^2$; então $A-B=8x$ e

$A+B=2(x^2+4+y^2)$.

$√A - √B = C ⇒ C^2(C^2 - 2(A+B)) + (A-B)^2 = 0$

$⇒ 4(4 - 4(x^2+4+y^2)) + (8x)^2 = 0$

$⇒ (-16x^2 - 16y^2 - 64 + 16) + 64x^2 = 0$

$⇒ 48x^2 - 16y^2 - 48 = 0$

\msk

$\begin{array}{ccc}
 (x,y) & d((x,y),(-2,0))-d((x,y),(2,0))=0 & 48x^2 - 16y^2 - 48=0 \\\hline
 (1,0) & 3-1=2 & 48-0-48 = 0 \\
 (2,3) & 5-3=2 & 192-144-48=0 \\
 (-2,3) & 5-3=2 & 192-144-48=0 \\
 (-1,0) & 1-2≠-2 & 48-0-48=0 \\
 \end{array}
$





%  ____  
% |___ \ 
%   __) |
%  / __/ 
% |_____|
%        


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

% (find-LATEX "edrxtikz.lua" "Line")

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



%  ____            ____   _ 
% |___ \ __ _     |___ \ (_)
%   __) / _` |_____ __) || |
%  / __/ (_| |_____/ __/ | |
% |_____\__,_|    |_____|/ |
%                      |__/ 
%
% «gab-2a-2j» (to ".gab-2a-2j")

\unitlength=7.5pt
\def\closeddot{\circle*{0.4}}

2a)                                  % a) x^2+y^2=1
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictellipse{v(0,0), v(1,0), v(0,1)}
   % \pictline{v(0,0), v(1,0), -2, 2}
   % \pictline{v(0,0), v(0,1), -2, 2}
   \end{picture}%
 }}$
\;
2b)                                  % b) x^2+y^2=4
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictellipse{v(0,0), v(2,0), v(0,2)}
   %\pictline{v(0,0), v(1,1), -2, 2}
   %\pictline{v(0,0), v(1,-1), -2, 2}
   \end{picture}%
 }}$
\;
2c)                                  % c) x^2+y^2=0
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \put(0,0){\closeddot}
   %\pictline{v(0,1), v(1,0), -2, 2}
   %\pictline{v(0,2), v(1,0), -2, 2}
   \end{picture}%
 }}$
%
\quad
%
2d)                                  % d) xy=1
$\vcenter{\hbox{%
   \beginpicture(-4,-4)(4,4)%
   \pictaxes%
   \picthyperbole{v(0,0), v(1,0), v(0,1), 1}{10, -4, -1/4, 1/4, 4}
   \put(1,1){\closeddot}
   \put(-1,-1){\closeddot}
   %\pictline{v(0,0), v(-1,1), -2, 2}
   \end{picture}%
 }}$
\;
2e)                                 % e) xy=4
$\vcenter{\hbox{%
   \beginpicture(-4,-4)(4,4)%
   \pictaxes%
   \picthyperbole{v(0,0), v(2,0), v(0,2), 1}{10, -2, -1/2, 1/2, 2}
   \put(2,2){\closeddot}
   \put(-2,-2){\closeddot}
   %\pictline{v(0,0), v(-1,1), -2, 2}
   \end{picture}%
 }}$
\;
2f)                                % f) xy=0
$\vcenter{\hbox{%
   \beginpicture(-3,-3)(3,3)%
   \pictaxes%
   \pictline{v(0,0), v(1,0), -3, 3}
   \pictline{v(0,0), v(0,1), -3, 3}
   \end{picture}%
 }}$
\;
2g)                               % g) xy=-1
$\vcenter{\hbox{%
   \beginpicture(-4,-4)(4,4)%
   \pictaxes%
   \picthyperbole{v(0,0), v(1,0), v(0,-1), 1}{10, -4, -1/4, 1/4, 4}
   \put(1,-1){\closeddot}
   \put(-1,1){\closeddot}
   \end{picture}%
 }}$
\;
2h)                               % h) x^2=y
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictparabola{v(0,0), v(1,0), v(0,1), 2}{10, -1.4, 1.4}
   \end{picture}%
 }}$
\quad
2i)                               % i) x=y^2
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictparabola{v(0,0), v(0,1), v(1,0), 2}{10, -1.4, 1.4}
   \end{picture}%
 }}$
\quad
2j)                               % j) x^2=y^2
$\vcenter{\hbox{%
   \beginpicture(-2,-2)(2,2)%
   \pictaxes%
   \pictline{v(0,0),  v(1,1), -2, 2}
   \pictline{v(0,0),  v(1,-1), -2, 2}
   \end{picture}%
 }}$


%  ____  _        ____      
% |___ \| | __   |___ \ ___ 
%   __) | |/ /____ __) / __|
%  / __/|   <_____/ __/\__ \
% |_____|_|\_\   |_____|___/
%                           
% «gab-2k-2s» (to ".gab-2k-2s")

\unitlength=7.5pt
\def\closeddot{\circle*{0.4}}

2k)                                   % k) u(u-1)=0
$\vcenter{\hbox{%
   \beginpicture(-3,-2)(3,2)%
   \pictaxes%
   \mygrid
   \pictline{v(0, 0), v(.5,-.5), -3, 3}
   \pictline{v(0, 1), v(.5,-.5), -1, 4}
   \end{picture}%
 }}$
\;
2l)                                   % l) v(v-1)=0
$\vcenter{\hbox{%
   \beginpicture(-3,-2)(3,2)%
   \pictaxes%
   \mygrid
   \pictline{v(0, 0),  v(1,0), -2,   2}
   \pictline{v(0,.5),  v(1,0), -2.5, 1.5}
   \end{picture}%
 }}$
\;
2m)                                   % m) u^2+v^2=0
$\vcenter{\hbox{%
   \beginpicture(-3,-2)(3,2)%
   \pictaxes%
   \mygrid
   \put(0,0){\closeddot}
   \end{picture}%
 }}$
\;
2n)                                   % n) u^2+v^2=1
$\vcenter{\hbox{%
   \beginpicture(-3,-2)(3,2)%
   \pictaxes%
   \mygrid
   \pictellipse{v(0,0), v(1,0), v(.5,-.5)}
   % \picthyperbole{v(2,-2), v(1,0), v(-1,1), 1}{10, -4, -1/2, 1/4, 4}
   % \put(2,-2){\closeddot}
   % \pictline{v(0,-2), v(1,0), -1, 5}  % y-2 = 0
   % \pictline{v(0, 0), v(1,-1), -2, 4}  % x+y = 0
   \end{picture}%
 }}$
\;
2o)                                    % o) uv=0
$\vcenter{\hbox{%
   \beginpicture(-3,-2)(3,2)%
   \pictaxes%
   \mygrid
   \pictline{v(0,0),  v(1,0), -3, 3}
   \pictline{v(0,0),  v(.5,-.5), -3, 3}
   \end{picture}%
 }}$
\quad
2p)                                    % p) uv=1
$\vcenter{\hbox{%
   \beginpicture(-3,-2)(3,2)%
   \pictaxes%
   \mygrid
   \picthyperbole{v(0,0), v(1,0), v(-.5,.5), 1}{10, -3, -1/3, 1/3, 3}
   \end{picture}%
 }}$
\;
2q)                                    % q) uv=-1
$\vcenter{\hbox{%
   \beginpicture(-4,-2)(4,2)%
   \pictaxes%
   \mygrid
   \picthyperbole{v(0,0), v(1,0), v(.5,-.5), 1}{10, -3, -1/3, 1/3, 3}
   \end{picture}%
 }}$
\;
2r)                                    % r) u^2=v
$\vcenter{\hbox{%
   \beginpicture(-4,-2)(4,2)%
   \pictaxes%
   \mygrid
   \pictparabola{v(0,0), v(1,0), v(-.5,.5), 2}{10, -1.4, 1.4}
   \end{picture}%
 }}$
\;
2s)
$\vcenter{\hbox{%
   \beginpicture(-4,-2)(4,2)%
   \pictaxes%
   \mygrid
   \pictparabola{v(0,0), v(-.5,.5), v(1,0), 2}{10, -1.4, 1.4}
   % \pictellipse{v(4,6), v(2,0), v(0,3)}
   \end{picture}%
 }}$


\bsk

%  _____ 
% |___ / 
%   |_ \ 
%  ___) |
% |____/ 
%        
% «gab-3» (to ".gab-3")

3) $z=\frac{3-x}{2}$, $y = 3-x-2 = 3 - x - \frac{3-x}{2} =
\frac{3-x}{2}$, $r=\setofst{(x, \frac{3-x}{2} \frac{3-x}{2})}{x∈\R}$

\bsk

%  _  _   
% | || |  
% | || |_ 
% |__   _|
%    |_|  
%         
% «gab-4» (to ".gab-4")

4) Sejam:

$\begin{array}{lll}
  r  = \setofst{(0,0,0)+\VEC{1,1,0}}{t∈\R},  & A=(0,0,0),  & \uu=\VEC{1,1,0}, \\
  s  = \setofst{(0,0,4)+\VEC{2,-2,0}}{t∈\R}, & B=(0,0,4),  & \vv=\VEC{2,-2,0}, \\
  r' = \setofst{(0,0,0)+\VEC{1,1,0}}{t∈\R},  & A'=(0,0,0), & \uu'=\VEC{1,1,0}, \\
  s' = \setofst{(0,0,3)+\VEC{1,0,-1}}{t∈\R}, & B'=(0,0,3), & \vv'=\VEC{1,0,-1}, \\
  \end{array}
$

4a) 4

4b) $d(r,s) = |([\uu,\vv,\Vec{AB}]/\area(\uu,\vv))| = |(16/||\VEC{0,0,4}||)| = 4$

4c) $d(r',s') = |([\uu',\vv',\Vec{A'B'}]/\area(\uu',\vv'))| = |(\vsm{1
  & 1 & 0 \\ 1 & 0 & -1 \\ 0 & 0 & 3 } / ||\VEC{1,1,-1}||)| =
|3/\sqrt3| = \sqrt3$




\end{document}

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