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Question Number 61864 by Tawa1 last updated on 10/Jun/19

Commented by MJS last updated on 10/Jun/19

I cannot read the fraction. Is it (3/2)k?

$$\mathrm{I}\:\mathrm{cannot}\:\mathrm{read}\:\mathrm{the}\:\mathrm{fraction}.\:\mathrm{Is}\:\mathrm{it}\:\frac{\mathrm{3}}{\mathrm{2}}{k}? \\ $$

Commented by Tawa1 last updated on 10/Jun/19

Yes sir

$$\mathrm{Yes}\:\mathrm{sir} \\ $$

Commented by MJS last updated on 10/Jun/19

ok I′ll solve it after dinner

$$\mathrm{ok}\:\mathrm{I}'\mathrm{ll}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{after}\:\mathrm{dinner} \\ $$

Answered by MJS last updated on 10/Jun/19

par: y=−x^2 +bx+c ((x),(y) )= ((0),((8k)) )∈par ⇒ c=8k ((x),(y) )= ((k),((6k)) )∈par ⇒ 6k=−k^2 +bk+8k ⇒ b=k−2 par: y=−x^2 +(k−2)x+8k tangent: y=αx+β method 1 {: (( ((x),(y) )= ((k),((6k)) )∈t ⇒ 6k=αk+β)),(( ((x),(y) )= ((((3k)/2)),(0) )∈t ⇒ 0=α((3k)/2)+β)) } ⇒ α=−12∧β=18k t: y=−12x+18k method 2 tangent of parabola y=−x^2 +bx+c in ((p),((−p^2 +bp+c)) ): t: y=(b−2p)x+(c+p^2 ) in our case t: y=−(k+2)x+k(k+8) t: y=−12x+18k ∧ t: y=−(k+2)x+k(k+8) ⇒ ⇒ −12=−(k+2) ∧ 18k=k(k+8) ⇒ k=10 par: y=−x^2 +8x+80 t: y=−12x+180 shaded area = ∫_k ^(3k/2) tdx−∫_k ^(3k/2) pardx= =∫_(10) ^(15) (x^2 −20x+100)dx=((125)/3)

$$\mathrm{par}:\:{y}=−{x}^{\mathrm{2}} +{bx}+{c} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{8}{k}}\end{pmatrix}\in\mathrm{par}\:\:\Rightarrow\:{c}=\mathrm{8}{k} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{{k}}\\{\mathrm{6}{k}}\end{pmatrix}\in\mathrm{par}\:\Rightarrow\:\mathrm{6}{k}=−{k}^{\mathrm{2}} +{bk}+\mathrm{8}{k}\:\Rightarrow\:{b}={k}−\mathrm{2} \\ $$$$\mathrm{par}:\:{y}=−{x}^{\mathrm{2}} +\left({k}−\mathrm{2}\right){x}+\mathrm{8}{k} \\ $$$$ \\ $$$$\mathrm{tangent}:\:{y}=\alpha{x}+\beta \\ $$$$\mathrm{method}\:\mathrm{1} \\ $$$$\left.\begin{matrix}{\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{{k}}\\{\mathrm{6}{k}}\end{pmatrix}\in\mathrm{t}\:\Rightarrow\:\mathrm{6}{k}=\alpha{k}+\beta}\\{\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\frac{\mathrm{3}{k}}{\mathrm{2}}}\\{\mathrm{0}}\end{pmatrix}\in\mathrm{t}\:\Rightarrow\:\mathrm{0}=\alpha\frac{\mathrm{3}{k}}{\mathrm{2}}+\beta}\end{matrix}\right\}\:\Rightarrow\:\alpha=−\mathrm{12}\wedge\beta=\mathrm{18}{k} \\ $$$$\mathrm{t}:\:{y}=−\mathrm{12}{x}+\mathrm{18}{k} \\ $$$$ \\ $$$$\mathrm{method}\:\mathrm{2} \\ $$$$\mathrm{tangent}\:\mathrm{of}\:\mathrm{parabola}\:{y}=−{x}^{\mathrm{2}} +{bx}+{c}\:\mathrm{in}\:\begin{pmatrix}{{p}}\\{−{p}^{\mathrm{2}} +{bp}+{c}}\end{pmatrix}:\: \\ $$$$\mathrm{t}:\:{y}=\left({b}−\mathrm{2}{p}\right){x}+\left({c}+{p}^{\mathrm{2}} \right) \\ $$$$\mathrm{in}\:\mathrm{our}\:\mathrm{case} \\ $$$$\mathrm{t}:\:{y}=−\left({k}+\mathrm{2}\right){x}+{k}\left({k}+\mathrm{8}\right) \\ $$$$ \\ $$$$\mathrm{t}:\:{y}=−\mathrm{12}{x}+\mathrm{18}{k}\:\wedge\:\mathrm{t}:\:{y}=−\left({k}+\mathrm{2}\right){x}+{k}\left({k}+\mathrm{8}\right)\:\Rightarrow \\ $$$$\Rightarrow\:−\mathrm{12}=−\left({k}+\mathrm{2}\right)\:\wedge\:\mathrm{18}{k}={k}\left({k}+\mathrm{8}\right)\:\Rightarrow\:{k}=\mathrm{10} \\ $$$$ \\ $$$$\mathrm{par}:\:{y}=−{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{80} \\ $$$$\mathrm{t}:\:\:\:\:\:\:{y}=−\mathrm{12}{x}+\mathrm{180} \\ $$$$ \\ $$$$\mathrm{shaded}\:\mathrm{area}\:=\:\underset{{k}} {\overset{\mathrm{3}{k}/\mathrm{2}} {\int}}\mathrm{t}{dx}−\underset{{k}} {\overset{\mathrm{3}{k}/\mathrm{2}} {\int}}\mathrm{par}{dx}= \\ $$$$=\underset{\mathrm{10}} {\overset{\mathrm{15}} {\int}}\left({x}^{\mathrm{2}} −\mathrm{20}{x}+\mathrm{100}\right){dx}=\frac{\mathrm{125}}{\mathrm{3}} \\ $$

Commented by Tawa1 last updated on 10/Jun/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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