71. (Mackenzie) Na figura adiante, as retas r e s são paralelas e a reta s é tangente à parábola de .... Pergunta de ideia deTensaigyn

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$\cdot\ Olhar\ anexo
\\\\Nota\c{c}\tilde{o}es\ : \\\\ \cdot\ \Rightarrow \ : \ essa\ representa\ ent\tilde{a}o\\\\ \cdot\ \Leftrightarrow \ : \ essa\ seta\ representa\ equivalente$

$\cdot\ K\ (\ x\ ,\ y\ )\ representa\ um\ ponto\ de\ abcissa\ x\ e\ ordenada\ y \\\\ \cdot \ r \ e\ s \ representam\ retas\\\\ \cdot\ \lambda\ representa\ uma\ par\acute{a}bola\ \\\\ \cdot\ x_1\ e\ x_2\ s\tilde{a}o\ as\ abcissas\ das\ ra\acute{i}zes\ da\ par\acute{a}bola\\\\ \cdot\ r\ //\ s\ indica\ o\ paralelismo\ entre\ as\ duas\ retas\\\\ \cdot\ m\ :\ coeficiente\ angular\ de\ uma\ reta\\\\ \cdot\ n\ :\ coeficiente\ linear\ de\ uma\ reta\\\\ \cdot\ \boxed{y'\ \acute{e}\ a\ derivada\ primeira\ de\ y}$

$--------\\\\ Propriedades\ e\ f\acute{o}rmulas\ adotadas\ :\\\\ \cdot\ \theta\ =\ arc\ tg\ x\ \ \Rightarrow\ \ tg\ \theta\ =\ x\\\\ \cdot\ Forma\ fatorada\ de\ uma\ par\acute{a}bola\ :\ y\ =\ a.(\ x\ -\ x_1\ ).(\ x\ -\ x_2\ )\\\\ \cdot\ Equa\c{c}\tilde{a}o\ reduzida\ da\ reta\ :\ y\ =\ m\ .\ x \ +\ n\\\\ 
\cdot Equa\c{c}\tilde{a}o\ geral\ da\ reta\ :\ a\ .\ x\ +\ b\ .\ y\ +\ n\ =\ 0\\\\ \cdot\ \boxed{\boxed{\ y'\ =\ m\ }}\\\\$

$\cdot\ Dist\hat{a}ncia\ entre\ duas\ retas\ paralelas\ :\ d\ =\ \frac{|\ n_1\ -\ n_2\ |}{\sqrt{a^2\ +\ b^2}}\\\\\\
--------$

$
Sabendo\ que\ P\ (\ -\ \sqrt{2}\ ,\ 0\ )\ \acute{e}\ uma\ das\ ra\acute{i}zes\ de\ \lambda\ e\ pelo\ fato\\\\ de\ o\ eixo\ de\ simetria\ par\acute{a}bola\ ser\ centrado\ na\ origem\ ent\tilde{a}o\\\\ podemos\ afirmar\ que\ Q\ (\ \sqrt{2}\ ,\ 0\ )\ \acute{e}\ a\ outra\ raiz\ .\ Se\ P\ e\ Q\\\\ s\tilde{a}o\ as\ ra\acute{i}zes\ da\ equa\c{c}\tilde{a}o\ ent\tilde{a}o\ x_1\ =\ -\ \sqrt{2}\ \ e\ \ x_2\ =\ \sqrt{2}\ .\\\\ Assim\ utilizando\ a\ forma\ fatorada\ da\ par\acute{a}bola\ :$

$y\ =\ a\ .\ (\ x\ -\ x_1\ )\ .\ (\ x\ -\ x_2\ )\\\\
y\ =\ a\ .\ [\ x\ -\ (\ - \ \sqrt{2})\ ]\ .\ (\ x\ -\ \sqrt{2}\ )\\\\
y\ =\ a\ .\ (\ x^2\ -\ 2\ )\\\\\\
Como\ \ V\ (\ 0\ ,\ -2\ )\ pertence\ a\ par\acute{a}bola\ ent\tilde{a}o\ vamos\ substituir\\\\ as\ coordenadas\ desse\ ponto\ na\ equa\c{c}\tilde{a}o\ logo\ acima\\\\\\
-2\ =\ a\ .\ [\ (\ 0\ )^2\ -\ 2\ ] \\\\
a\ =\ 1$

$Se\ \ a\ =\ 1\ \ ent\tilde{a}o\ vamos\ substitu\acute{i}-lo\ ,\\\\

y\ =\ a\ .\ (\ x^2\ -\ 2\ )\\\\
y\ =\ (1)\ .\ (\ x^2\ -\ 2\ )\\\\
y\ =\ x^2\ -\ 2\\\\
Assim\ temos\ que\ ,\\\\\
\boxed{\boxed{\lambda\ : \ y\ =\ x^2\ -\ 2}}$

$Podemos\ afirmar\ que\ :\\\\
\alpha\ =\ arc\ tg\ 2\ \ \Rightarrow\ \ tg\ \alpha\ =\ 2\\\\
E\ que\ ,\\\\
m_r \ =\ tg\ \alpha\ \ \Rightarrow\ \ m_r\ =\ 2\\\\ 
Utilizando\ a\ equa\c{c}\tilde{a}o\ reduzida\ da\ reta\ ,\\\\
y\ =\ m_r\ .\ x\ +\ n_r\\\\
y\ =\ 2\ .\ x\ +\ n_r \\\\
Analisando\ o\ gr\acute{a}fico\ temos\ que\ o\ ponto\ T\ (\ 6\ ,\ 0\ )\ pertence\ a\ r\ . \\\\ Vamos\ substituir\ essas\ coordenadas\ na\ equa\c{c}\tilde{a}o\ acima \ ,\\\\
0\ =\ 2\ .\ (6)\ +\ n_r\\\\
n_r\ =\ -12$

$Sabendo\ que\ \ n_r\ =\ -12\ \ vamos\ substitu\acute{i}-lo\ ,\\\\
y\ =\ 2\ .\ x\ +\ n_r\\\\
y\ =\ 2\ .\ x\ +\ (-12)\\\\
y\ =\ 2\ .\ x\ -\ 12\\\\
Assim\ temos\ que\ ,\\\\
\boxed{\boxed{r\ :\ y\ =\ 2\ .\ x\ -\ 12 \ \Leftrightarrow r\ :\ 2\ .\ x\ -\ y\ -\ 12\ =\ 0}}$

$Podemos\ afirmar\ tamb\acute{e}m\ ,\\\\
r\ //\ s\ \ \Rightarrow\ \ m_r\ =\ m_s\ \ \Rightarrow\ \ m_s\ =\ 2\\\\
Assim\ temos\ que\ ,\\\\
y\ =\ m_s\ .\ x\ +\ n_s\\\\
\boxed{\boxed{s\ :\ y\ =\ 2\ .\ x\ +\ n_s}}$

$Em\ c\acute{a}lculo\ temos\ que\ \ \boxed{\boxed{y'\ =\ m}}\ \ ,\ ou\ seja\ ,\ a\ derivada\\\\ primeira\ de\ uma\ fun\c{c}\tilde{a}o\ funciona\ como\ o\ coeficiente\ angular\\\\ da\ reta\ tangente\ a\ mesma\ no\ ponto\ B\ (x,y). \ Derivando\ \lambda\ , \\\\\\
y\ =\ x^2\ -\ 2 \ \ \ \ \Rightarrow\ \ \ \ \ \ y' \ =\ 2x\\\\\\
Como\ sabemos\ que\ a\ reta\ s\ de\ coeficiente\ angular\ m_s\ =\ 2 \\\\ \acute{e}\ tangente\ \grave{a}\ \lambda\ ,\ ent\tilde{a}o :\\\\\\
y'\ =\ m_s\\\\
2x\ =\ 2\\\\
x\ =\ 1$

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