% (find-angg "LATEX/2017-1-GA-P2.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2017-1-GA-P2.tex"))
% (defun d () (interactive) (find-xpdfpage "~/LATEX/2017-1-GA-P2.pdf"))
% (defun e () (interactive) (find-LATEX "2017-1-GA-P2.tex"))
% (defun u () (interactive) (find-latex-upload-links "2017-1-GA-P2"))
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% (find-sh0 "cp -v  ~/LATEX/2017-1-GA-P2.pdf /tmp/pen/")
%   file:///home/edrx/LATEX/2017-1-GA-P2.pdf
%               file:///tmp/2017-1-GA-P2.pdf
%           file:///tmp/pen/2017-1-GA-P2.pdf
% http://angg.twu.net/LATEX/2017-1-GA-P2.pdf
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%
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%
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\def\pu{\directlua{pu()}}

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%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 - 2017.1
\par P2 - 17/jul/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 $ax^2 + bxy +
cy^2 + dy + ey + f = 0$; $4+(x+y)(x-y)=5y$ não é uma equação de cônica
mas é equivalente a uma: $x^2-y^2-5y+4=0$. 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

% \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}}

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

\bsk




2) \T(Total: 1.5 pts) Seja $S=\setofxyst{y-x/2 = (y+x/2)^2}$ (veja o
item $s$ da questão anterior). Dê as coordenadas $(x,y)$ de cinco
pontos de $S$.

{\sl Dica: confira os seus resultados!!!}

\bsk




3) \T(Total: 1.5 pts) Sejam $F=(-1,0)$, $F'=(1,0)$, e
%
$$\begin{array}{rcl}
  S  &=& \setofxyst{d((x,y),F) + d((x,y),F')=4} \\
     &=& \setofxyst{\sqrt{(x+1)^2+y^2} + \sqrt{(x-1)^2+y^2} = 4}, \\
  S' &=& \setofxyst{ax^2 + bx + c + dxy + ey + fy^2 = 0}. \\
  \end{array}
$$
%
a) \B(1.0 pts) Encontre $a,b,c,d,e,f∈\R$ que façam com que $S=S'$.

b) \B(0.5 pts) Dê as coordenadas de 4 pontos da elipse $S$.

\bsk




4) \T(Total: 2.5 pts) Sejam $A=(0,0,2)$, $B=(0,1,0)$, $C=(1,0,0)$,
$D=(1,1,0)$, $π$ o plano que contém $A,B,C$, $π'$ o plano paralelo a
$π$ que contém o ponto $D$.

a) \B(0.2 pts) Dê a equação do plano $π$.

b) \B(0.3 pts) Dê a equação do plano $π'$.

c) \B(0.5 pts) Calcule $d(π,π')$.

d) \B(1.5 pts) Dê as coordenadas do ponto de $π'$ mais próximo de $A$.





\newpage

{\bf Gabarito} (versão revisada)

\msk

%  _ 
% / |
% | |
% | |
% |_|
%    

\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)}}}

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




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



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

\msk

1o)                                    % o) uv=0
$\vcenter{\hbox{%
   \beginpicture(-4,-2)(4,2)%
   \pictaxes%
   \mygrid
   \pictline{v(0,0),  v(1,.5), -3, 3}
   \pictline{v(0,0),  v(1,-.5), -3, 3}
   \end{picture}%
 }}$
\quad
1p)                                    % p) uv=1
$\vcenter{\hbox{%
   \beginpicture(-4,-2)(4,2)%
   \pictaxes%
   \mygrid
   \picthyperbole{v(0,0), v(1,.5), v(-1,.5), 1}{10, -3, -1/3, 1/3, 3}
   \end{picture}%
 }}$
\;
1q)                                    % q) uv=-1
$\vcenter{\hbox{%
   \beginpicture(-4,-2)(4,2)%
   \pictaxes%
   \mygrid
   \picthyperbole{v(0,0), v(1,.5), v(1,-.5), 1}{10, -3, -1/3, 1/3, 3}
   \end{picture}%
 }}$
\;
1r)                                    % r) u^2=v
$\vcenter{\hbox{%
   \beginpicture(-4,-2)(4,2)%
   \pictaxes%
   \mygrid
   \pictparabola{v(0,0), v(-1,.5), v(1,.5), 2}{10, -1.4, 1.4}
   \end{picture}%
 }}$
\;
1s)
$\vcenter{\hbox{%
   \beginpicture(-4,-2)(4,2)%
   \pictaxes%
   \mygrid
   \pictparabola{v(0,0), v(1,.5), v(-1,.5), 2}{10, -1.4, 1.4}
   % \pictellipse{v(4,6), v(2,0), v(0,3)}
   \end{picture}%
 }}$




\bsk

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

2)
% $\begin{array}[t]{rrrrl}
%  u & v & x & y \\\hline
%  4 & -2 & -6 & 1 & \text{(0.5 pts)} \\
%  1 & -1 & -2 & 0 & \text{(0.2 pts)} \\
%  0 &  0 &  0 & 0 & \text{(0.1 pts)} \\
%  1 &  1 &  0 & 1 & \text{(0.2 pts)} \\
%  4 &  2 & -2 & 3 & \text{(0.5 pts)} \\
%  \end{array}
% $
% 
$\begin{array}[t]{rrrrl}
 u & v & x & y \\\hline
 4 & -2 & -6 & 1 & \text{(0.5 pts)} \\
 1 & -1 & -2 & 0 & \text{(0.2 pts)} \\
 0 &  0 &  0 & 0 & \text{(0.1 pts)} \\
 1 &  1 &  0 & 1 & \text{(0.2 pts)} \\
 4 &  2 & -2 & 3 & \text{(0.5 pts)} \\
 \end{array}
$

\bsk




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

3a) Sejam
%
$$\begin{array}{rcl}
  A &=& (x+1)^2+y^2 = x^2 + 2x + 1 + y^2, \\
  B &=& (x-1)^2+y^2 = x^2 - 2x + 1 + y^2, \\
  C &=& 4. \\
  \end{array}
$$
%
Então
%
$$\begin{array}{rcl}
  A+B &=& 2x^2 + 2y^2 + 2, \\
  A-B &=& 4x, \\
  C^2 - 2(A+B) &=& 16 - 4x^2 - 4y^2 - 4 = 12 - 4x^2 - 4y^2, \\
  C^2(C^2 - 2(A+B)) &=& 192 - 64x^2 - 64y^2, \\
  (A-B)^2 &=& 16x^2, \\
  C^2(C^2 - 2(A+B)) + (A-B)^2 &=& 192 - 48x^2 - 64y^2 \\
                              &=& 192(1 - \frac{1}{4}{x^2} - \frac{1}{3}{y^2}), \\
  \end{array}
$$
%
$$\begin{array}{rcl}
  S' &=& \setofxyst{-48x^2 +0x +192 +0xy + 0y - 64y^2 = 0} \\
     &=& \setofxyst{\frac{1}{4}{x^2} + 0x - 1 +0xy + 0y +\frac{1}{3}{y^2} = 0}. \\
  \end{array}
$$

3b) $\begin{array}[t]{rcl}
             & (0,√3), &       \\
     (-2,0), &        & (2,0), \\
             & (0,-√3). &       \\
     \end{array}
    $

% (find-es "ipython" "2017.1-GA-P2")


\bsk

%  _  _   
% | || |  
% | || |_ 
% |__   _|
%    |_|  
%         

4a) $π=\setofxyzst{x+y+\frac z2 = 1}$

4b) $π'=\setofxyzst{x+y+\frac z2 = 2}$

4c) Sejam $\uu=\Vec{AB}=\VEC{0,1,-2}$, $\vv=\Vec{AC}=\VEC{1,0,-2}$,
$\ww=\Vec{AD}=\VEC{1,1,-2}$. Então
%
$$\begin{array}{rcl}
 \uu×\vv &=& \VEC{-2,-2,-1}, \\
 \area(\uu,\vv) &=& \sqrt{2^2 + 2^2 + 1^2}
                = 3, \\
 |[\uu,\vv,\ww]| &=& \vsm{0 & 1 & -2 \\ 1 & 0 & -2 \\ 1 & 1 & -2}
                 = -2, \\
 d(π,π') &=& \left| \frac{|[\uu,\vv,\ww]|}{\area(\uu,\vv)} \right|
          = |-2/3|
          = 2/3.
 \end{array}
$$

4d) A reta $r = \setofst{A+t\VEC{2,2,1}}{t∈\R} =
\setofst{(2t,2t,2+t)}{t∈\R}$ passa por $A$ e é ortogonal a $π$. Ela
intersecta $π'$ quando $2t+2t+\frac{2+t}{2}=2$, i.e., quando $\frac92
t=1$ e quando $t=\frac29$, o que corresponde ao ponto $P=(\frac49,
\frac49, \frac{20}{9})$.


% (find-angg "LATEX/2015-2-GA-P2.tex")

% n = (2,2,1)
% u = (1,0,-2)
% u = (0,1,-2)
% 
% A=(0,0,2)
% B=(0,1,0)
% C=(1,0,0)
% D=(1,1,0)




\end{document}

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