Warning: this is an htmlized version!
The original is here, and
the conversion rules are here.
% (find-angg "LATEX/2015-2-GA-VR.tex")
% (find-angg "LATEX/2015-2-GA-VR.lua")
% (defun c () (interactive) (find-LATEXsh "lualatex 2015-2-GA-VR.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2015-2-GA-VR.pdf"))
% (defun e () (interactive) (find-LATEX "2015-2-GA-VR.tex"))
% (defun u () (interactive) (find-latex-upload-links "2015-2-GA-VR"))
% (defun z () (interactive) (find-zsh "flsfiles-tgz 2015-2-GA-VR.fls 2015-2-GA-VR.tgz")
% (find-xpdfpage "~/LATEX/2015-2-GA-VR.pdf")
% (find-sh0 "cp -v  ~/LATEX/2015-2-GA-VR.pdf /tmp/")
% (find-sh0 "cp -v  ~/LATEX/2015-2-GA-VR.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2015-2-GA-VR.pdf
%               file:///tmp/2015-2-GA-VR.pdf
%           file:///tmp/pen/2015-2-GA-VR.pdf
% http://angg.twu.net/LATEX/2015-2-GA-VR.pdf
\documentclass[oneside]{book}
\usepackage[colorlinks]{hyperref} % (find-es "tex" "hyperref")
%\usepackage[latin1]{inputenc}
\usepackage{amsmath}
\usepackage{amsfonts}
\usepackage{amssymb}
\usepackage{tikz}
%
\usepackage{edrx15}               % (find-angg "LATEX/edrx15.sty")
\input edrxaccents.tex            % (find-angg "LATEX/edrxaccents.tex")
\input edrxheadfoot.tex           % (find-dn4ex "edrxheadfoot.tex")
\input istanbuldefs               % (find-LATEX "istanbuldefs.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")
%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end

\def\setofet  #1{\setofst{#1}{t∈\R}}

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

{\setlength{\parindent}{0em}
\footnotesize
\par Geometria Analítica
\par PURO-UFF - 2015.2
\par VR - 28/mar/2016 - Eduardo Ochs
\par Links importantes:
\par \url{http://angg.twu.net/2015.2-GA.html} (página do curso)
\par \url{http://angg.twu.net/2015.2-GA/2015.2-GA.pdf} (quadros)
\par \url{http://angg.twu.net/LATEX/2015-2-GA-VR.pdf} (esta prova)
%\par \url{http://angg.twu.net/LATEX/2015-2-GA-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)}}


1) \T(Total: 3.5 pts) Sejam $r:y=\frac34x-2$ e $s:y=2x-7$.

a) \B(0.5 pts) Seja $A∈r∩s$. Dê as coordenadas de $A$.

b) \B(0.5 pts) Encontre um ponto $P∈s$ tal que $d(P,r)=2$.

c) \B(0.5 pts) Dê a equação do círculo $C$ centrado em $P$ que passa por $A$.

d) \B(0.5 pts) Encontre o ponto $B∈r$ mais próximo de $P$.

e) \B(0.5 pts) Dê as coordenadas dos dois pontos de $C∩r$.

f) \B(0.5 pts) Dê as coordenadas dos dois pontos de $C∩s$.

g) \B(0.5 pts) Represente tudo graficamente.

\bsk

2) \T(Total: 1.5 pts) Sejam $E:16x^2 - 160x + 25y^2 - 300y + 900 = 0$.

a) \B(0.5 pts) Dê a equação reduzida da elipse $E$.

b) \B(0.5 pts) Encontre quatro pontos $P_0, P_1, P_2, P_3 ∈ E$.

c) \B(0.5 pts) Encontre os focos $F_1$ e $F_2$ de $E$.

\bsk

3) \T(Total: 2.0 pts) Sejam $A=(6,0,0)$, $B=(0,6,0)$, $C=(0,0,6)$,
$D=(2,4,5)$, $P=(4,4,4)$, $r=\setofet{A+\Vec{AB}}$, $π'$ o plano que
contém $r$ e $C$, e $π''$ o plano que contém $r$ e $D$.

a) \B(0.5 pts) Encontre o ponto $Q∈r$ mais próximo de $P$.

b) \B(0.5 pts) Encontre o ponto $P'∈π'$ mais próximo de $P$.

c) \B(0.5 pts) Encontre o ponto $P''∈π''$ mais próximo de $P$.

d) \B(1.0 pts) Mostre que $P$, $P'$, $P''$ e $Q$ são coplanares.

\bsk

4) \T(Total: 2.5 pts) Sejam $A=(4,0,0)$, $B=(0,4,0)$, $C=(0,0,7)$,
$D=(0,7,0)$. Sejam $r$ uma reta que passa por $A$ e $B$, $r'$ uma reta
que passa por $C$ e $D$, e $s$ uma reta ortogonal a $r$ e $r'$ que
corta ambas. Dê uma parametrização para $s$.




% \newpage

%  _   _ _         
% | |_(_) | __ ____
% | __| | |/ /|_  /
% | |_| |   <  / / 
%  \__|_|_|\_\/___|
%                  

% Dots, labels, vectors
%
\def\uu{\vec u}
\def\vv{\vec v}
\def\ww{\vec w}
\def\nn{\vec n}
\def\VEC#1{{\overrightarrow{(#1)}}}

\def\nm#1{\|#1\|}
\def\Reg#1{(#1)}

\def\setofxyst #1{\setofst{(x,y)  ∈\R^2}{#1}}
\def\setofxyzst#1{\setofst{(x,y,z)∈\R^3}{#1}}
\def\setofet  #1{\setofst{#1}{t∈\R}}
\def\setofeu  #1{\setofst{#1}{u∈\R}}
\def\setofpt  #1 #2 #3 #4 {\setofet{(#1,#2)+t\VEC{#3,#4}}}
\def\setofpu  #1 #2 #3 #4 {\setofeu{(#1,#2)+u\VEC{#3,#4}}}

\def\setofeaa  #1{\setofst{#1}{α∈\R}}
\def\setofebb  #1{\setofst{#1}{β∈\R}}

% \mygrid and \myaxes
% (find-es "tikz" "mygrid")
\tikzset{mycurve/.style=very thick}
\tikzset{axis/.style=semithick}
\tikzset{tick/.style=semithick}
\tikzset{grid/.style=gray!20,very thin}
\tikzset{anydot/.style={circle,inner sep=0pt,minimum size=1.2mm}}
\tikzset{opdot/.style={anydot, draw=black,fill=white}}
\tikzset{cldot/.style={anydot, draw=black,fill=black}}
%
\def\mygrid(#1,#2) (#3,#4){
  \clip              (#1-0.4, #2-0.4) rectangle (#3+0.4, #4+0.4);
  \draw[step=1,grid] (#1-0.2, #2-0.2) grid      (#3+0.2, #4+0.2);
  \draw[axis] (-10,0) -- (10,0);
  \draw[axis] (0,-10) -- (0,10);
  \foreach \x in {-10,...,10} \draw[tick] (\x,-0.2) -- (\x,0.2);
  \foreach \y in {-10,...,10} \draw[tick] (-0.2,\y) -- (0.2,\y);
}
\def\myaxes(#1,#2) (#3,#4){
  \clip              (#1-0.4, #2-0.4) rectangle (#3+0.4, #4+0.4);
 %\draw[step=1,grid] (#1-0.2, #2-0.2) grid      (#3+0.2, #4+0.2);
  \draw[axis] (-30,0) -- (30,0);
  \draw[axis] (0,-30) -- (0,30);
  \foreach \x in {-30,...,30} \draw[tick] (\x,-0.2) -- (\x,0.2);
  \foreach \y in {-30,...,30} \draw[tick] (-0.2,\y) -- (0.2,\y);
}

% Grid color
\tikzset{grid/.style=gray!50,very thin}

\def\tikzp#1{\mat{\begin{tikzpicture}#1\end{tikzpicture}}}

\def\mydraw       #1;{\draw [mycurve]  \expr{#1};}
\def\mydot        #1;{\node [cldot] at \expr{#1} [] {};}
\def\myldot #1 #2 #3;{\node [cldot] at \expr{#1} [label=#2:${#3}$] {};}
\def\myseg     #1 #2;{\draw [mycurve]  \expr{#1} -- \expr{#2};}
\def\mylabel #1 #2 #3;{\node []     at \expr{#1} [label=#2:${#3}$] {};}
\def\myseggrid  #1 #2;{\draw [grid]    \expr{#1} -- \expr{#2};}

\def\e{\expr}

% (find-dn6 "picture.lua" "V")
%L V.__tostring = function (v) return format("(%.3f,%.3f)", v[1], v[2]) end
%L V.__div      = function (v, k) return v*(1/k) end
%L V.__index.tow = function (A, B, t) return A+(B-A)*t   end  -- towards
%L V.__index.mid = function (A, B)    return A+(B-A)*0.5 end  -- midpoint

\newpage

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

Mini-gabarito:

\msk

1a) $A=(4,1)$

1b) $P = A+\VEC{2,4} = (6,5)$

1c) $C = \setofxyst{(x-6)^2 + (y-5)^2 = \sqrt{20}}$

1d) $B = A+\VEC{3.2, 2.4} = (7.2, 3.4)$

1e) $B' = A+2\Vec{AB}$; $C∩r = \{A, B'\}$

1f) $P' = A+2\Vec{AP}$; $C∩s = \{A, P'\}$

\msk
Tem uma segunda solução, com:

1b) $P = A-\VEC{2,4} = (2,-3)$

1c) $C = \setofxyst{(x-2)^2 + (y+3)^2 = \sqrt{20}}$

1d) $B = A-\VEC{3.2, 2.4} = (0.8, -1.4)$

Ela está desenhada abaixo à direita.

\msk

%L r = Line.new(v(0, -2), v(1, 3/4), -3, 12)
%L s = Line.new(v(0, -7), v(1, 2),   -3, 12)
%L A = v(4, 1)
%L P = A + v(2.0, 4.0)
%L B = A + v(3.2, 2.4)
%L PP = A:tow(P, 2)
%L BB = A:tow(B, 2)
%L C  = Ellipse.newcircle(P, math.sqrt(20))
\pu
$\tikzp{[scale=0.3,auto]
    \mygrid (-1,-7) (12,10);
    % \draw [mycurve]  \e{A} -- \e{B} -- \e{C} -- \e{D} -- \e{A};
    \mydraw r:draw();
    \mydraw s:draw();
    \mydraw C:draw();
    \mylabel r:t(1) 270 r;
    \mylabel s:t(1)   0 s;
    \myldot A 180 A;
    \myldot P 135 P;
    \myldot B 270 B;
    \myldot PP 0 P';
    \myldot BB 0 B';
    \myseg B P;
  }
$
\quad
%L r = Line.new(v(0, -2), v(1, 3/4), -3, 12)
%L s = Line.new(v(0, -7), v(1, 2),   -3, 12)
%L A = v(4, 1)
%L P = A - v(2.0, 4.0)
%L B = A - v(3.2, 2.4)
%L PP = A:tow(P, 2)
%L BB = A:tow(B, 2)
%L C  = Ellipse.newcircle(P, math.sqrt(20))
\pu
$\tikzp{[scale=0.3,auto]
    \mygrid (-5,-9) (8,6);
    % \draw [mycurve]  \e{A} -- \e{B} -- \e{C} -- \e{D} -- \e{A};
    \mydraw r:draw();
    \mydraw s:draw();
    \mydraw C:draw();
    \mylabel r:t(6) 270 r;
    \mylabel s:t(6)   0 s;
    \myldot A  90 A;
    \myldot P   0 P;
    \myldot B 135 B;
    \myldot PP 315 P';
    \myldot BB 135 B';
    \myseg B P;
  }
$


% Segunda solução:
% 1b) $P = A-\VEC{2,4} = (2,-3)$
% 1d) $B = A-\VEC{3.2, 2.4} = (2.8, 1.6)$



\bsk

2a) $(\frac{x-5}{5})^2 + (\frac{y-6}{4})^2 = 1$

2b) $(5-5,6)$, $(5+5,6)$, $(5,6+4)$, $(5,6-4)$

2c) $(5-3,6)$, $(5+3,6)$

\bsk

3) $π'=\setofxyzst{x+y+z=6}$, $π''=\setofxyzst{x+y=6}$

3a) $Q=(3,3,0)$

3b) $P'=(2,2,2)$

3c) $P''=(3,3,4)$

3d) $[\Vec{PP'}, \Vec{PP''}, \Vec{PQ}] = 0$ 

\bsk

4) $s = \setofet{(-1,5,0) + t\VEC{1,1,1}}$


\end{document}



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