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Warning: this is an htmlized version!
The original is here, and the conversion rules are here. |
% (find-angg "LATEX/2017-1-GA-P2.tex")
% (defun c () (interactive) (find-LATEXsh "lualatex -record 2017-1-GA-P2.tex"))
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% ____ _ _ _
% / ___|__ _| |__ ___ ___ __ _| | |__ ___
% | | / _` | '_ \ / _ \/ __/ _` | | '_ \ / _ \
% | |__| (_| | |_) | __/ (_| (_| | | | | | (_) |
% \____\__,_|_.__/ \___|\___\__,_|_|_| |_|\___/
%
{\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()}}}
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\vcenter{\hbox{%
\beginpicture(#1)(#2)%
\pictaxes%
\pictline{#3}
\end{picture}%
}}%
}
\def\pictellipse#1{{\linethickness{1.0pt}\expr{Ellipse.new(#1):pict()}}}
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\vcenter{\hbox{%
\beginpicture(#1)(#2)%
\pictaxes%
\pictellipse{#3}
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}}%
}
\def\pictEllipseF(#1)(#2)#3(#4)(#5){%
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\pictaxes%
\pictellipse{#3}
\put(#4){\closeddot}
\put(#5){\closeddot}
\end{picture}%
}}%
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\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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