Question

71. (Mackenzie) Na figura adiante, as retas r e s são

paralelas e a reta s é tangente à parábola de vértice

(0, -2). Então a distância d entre r e s é:

a) (7√5)/5

b) (8√5)/5

c) (9√5)/5

d) (11√5)/5

e) (12√5)/5


A resposta é a letra C.

Me ajudem, por favor. Estou sofrendo com essa questão!!

Answer 1

Fiz por desenho por achar mais didático, mas veja se compreendeu.

Answer 2

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

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

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

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

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

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

$Se\ \ a\ =\ 1\ \ ent\tilde{a}o\ vamos\ substitu\acute{i}-lo\ ,\\\\ y\ =\ a\ .\ (\ x^2\ -\ 2\ )\\\\ y\ =\ (1)\ .\ (\ x^2\ -\ 2\ )\\\\ y\ =\ x^2\ -\ 2\\\\ Assim\ temos\ que\ ,\\\\\ \boxed{\boxed{\lambda\ : \ y\ =\ x^2\ -\ 2}}$

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

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

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

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

$Se\ x\ =\ 1\ representa\ a\ abcissa\ do\ ponto\ B\ comum\ a\ s\ e\ \lambda \ ,\\\\ ent\tilde{a}o\ para\ acharmos\ a\ ordenada\ basta\ substituirmos\ na\\\\ equa\c{c}\tilde{a}o\ da\ par\acute{a}bola\ ,\\\\ y\ =\ x^2\ -\ 2\\\\ y\ =\ (1)^2\ -\ 2\\\\ y\ =\ -1\\\\ Sendo\ assim\ temos\ que\ B\ (\ 1\ ,\ -1\ )\ . \ Agora\ para\ descobrirmos\\\\ a\ equa\c{c}\tilde{a}o\ de\ s\ basta\ substituirmos\ as\ coordenadas\ de\ B\ ,\\\\ y\ =\ 2\ .\ x\ +\ n_s\\\\ -1\ =\ 2\ .\ (1)\ +\ n_s\\\\ n_s\ =\ -\ 3$

$Assim\ temos\ que\ :\\\\ \boxed{\boxed{s\ :\ y\ =\ 2\ .\ x\ -\ 3\ \ \ \ \Leftrightarrow\ \ \ \ s\ :\ 2\ .\ x\ -\ y\ -\ 3\ =\ 0}}\\\\\\\ Sabendo\ que\ r\ //\ s\ ent\tilde{a}o\ podemos\ calcular\ a\ dist\hat{a}ncia\ das\\\\ duas\ retas\ por\ :\\\\ d\ =\ \frac{| \ n_r \ -\ n_s|}{\sqrt{a^2\ +\ b^2}}\\\\ Temos\ que\ : \ \ a\ =\ 2\ \ ;\ \ b\ =\ -1\ \ ;\ \ n_r\ =\ -12\ \ ;\ \ n_s\ =\ -3\ \ .\\\\ Substituindo\ ,$

$d\ =\ \frac{|\ (-12)\ -\ (-3)\ |}{\sqrt{(2)^2\ +\ (-1)^2}}\\\\ d\ =\ \frac{|-9|}{\sqrt{5}}\\\\ d\ =\ \frac{9 \ .\ \sqrt{5}}{5} \ \ \ \ \ \ \ \ \ \ \ letra \ c)$