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% (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 Analítica
\par PURO-UFF - 2016.2
\par P2 - 18/jan/2017 - Eduardo Ochs
\par Respostas sem justificativas não serão aceitas.
\par Proibido usar quaisquer aparelhos eletrônicos.
% \par Versão: 14/mar/2016
% \par Links importantes:
% \par \url{http://angg.twu.net/2015.2-C2.html} (página 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 equação de cônica é uma equação da forma $ax^2 + bxy +
cy^2 + dy + ey + f = 0$; $4+(x+y)(x-y)=5y$ não é uma equação de cônica
mas é equivalente a uma: $x^2-y^2-5y+4=0$.
Nas questões 3 e 4 vamos usar a abreviação $[\text{equação}] =
\setofxyzst{\text{equação}}$.
\bsk
1) \T(Total: 4.0 pts) Faça esboços das cônicas com as equações 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 óbvios 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 $π_1=[y=3-2x]$, $π_2=[x+y+z=4]$, $r=π_1∩π_2$.
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 $r'=\setofst{(x,ax+b,cx+d)}{x∈\R}$ e $r=r'$, quem
são $a$, $b$, $c$ e $d$?
\bsk
\bsk
4) \T(Total: 3.5 pts) Sejam $A=(1,0,0)$, $B=(0,1,0)$, $\uu=\vec{AB}$,
$C=(1,1,0)$, $D=(1,0,1)$, $\vv=\vec{CD}$, $r=\setofst{A+t\uu}{t∈\R}$,
$r'=\setofst{C+t'\vv}{t'∈\R}$.
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 distância entre $r$ e $r'$.
\newpage
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% | | _ / _` | '_ \ / _` | '__| | __/ _ \
% | |_| | (_| | |_) | (_| | | | | || (_) |
% \____|\__,_|_.__/ \__,_|_| |_|\__\___/
%
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) $x=0 \;⇒\; y=3 \;⇒\; z=1 \;⇒\; P=(0,3,1)$
3b) $x=1 \;⇒\; y=1 \;⇒\; z=2 \;⇒\; P=(1,1,2)$
3c) $r'=\setofst{(x,-2x+3,1x+1)}{x∈\R}$, $a=-2$, $b=3$, $c=1$ 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
%
$$\begin{array}{rcl}
d(r,r') &=& [\Vec{AB}, \Vec{CD}, \Vec{AC}] / ||\Vec{AB} × \Vec{CD}|| \\
&=& [\V(-1,1,0), \V(0,-1,1), \V(0,1,0)] / ||\V(-1,1,0) × \V(0,-1,1)|| \\
&=& \vsm{-1&1&0 \\ 0&-1&1 \\ 0&1&0} / ||\V(1,1,1)|| \\
&=& 1/\sqrt{3} \\
&=& \sqrt{3}/3 \\
\end{array}
$$
%
\end{document}
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