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

Commented by MJS last updated on 10/Jun/19
$I cannot read the fraction. Is it (3/2)k?$
$I cannot read the fraction . Is it \frac{3}{2} k ?$
Commented by Tawa1 last updated on 10/Jun/19

$Yes sir$
Commented by MJS last updated on 10/Jun/19

$ok I^{'} ll solve it after dinner$
	Answered by MJS last updated on 10/Jun/19
	$par: y=−x^2 +bx+c ((x),(y) )= ((0),((8k)) )∈par ⇒ c=8k ((x),(y) )= ((k),((6k)) )∈par ⇒ 6k=−k^2 +bk+8k ⇒ b=k−2 par: y=−x^2 +(k−2)x+8k tangent: y=αx+β method 1 {: (( ((x),(y) )= ((k),((6k)) )∈t ⇒ 6k=αk+β)),(( ((x),(y) )= ((((3k)/2)),(0) )∈t ⇒ 0=α((3k)/2)+β)) } ⇒ α=−12∧β=18k t: y=−12x+18k method 2 tangent of parabola y=−x^2 +bx+c in ((p),((−p^2 +bp+c)) ): t: y=(b−2p)x+(c+p^2 ) in our case t: y=−(k+2)x+k(k+8) t: y=−12x+18k ∧ t: y=−(k+2)x+k(k+8) ⇒ ⇒ −12=−(k+2) ∧ 18k=k(k+8) ⇒ k=10 par: y=−x^2 +8x+80 t: y=−12x+180 shaded area = ∫_k ^(3k/2) tdx−∫_k ^(3k/2) pardx= =∫_(10) ^(15) (x^2 −20x+100)dx=((125)/3)$
	$par : y = - x^{2} + b x + c$ $(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} 0 \\ 8 k \end{matrix}) \in par \Rightarrow c = 8 k$ $(\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} k \\ 6 k \end{matrix}) \in par \Rightarrow 6 k = - k^{2} + b k + 8 k \Rightarrow b = k - 2$ $par : y = - x^{2} + (k - 2) x + 8 k$ $tangent : y = α x + β$ $method 1$ $\begin{matrix} (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} k \\ 6 k \end{matrix}) \in t \Rightarrow 6 k = α k + β \\ (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} \frac{3 k}{2} \\ 0 \end{matrix}) \in t \Rightarrow 0 = α \frac{3 k}{2} + β \end{matrix}} \Rightarrow α = - 12 \land β = 18 k$ $t : y = - 12 x + 18 k$ $method 2$ $tangent of parabola y = - x^{2} + b x + c in (\begin{matrix} p \\ - p^{2} + b p + c \end{matrix}) :$ $t : y = (b - 2 p) x + (c + p^{2})$ $in our case$ $t : y = - (k + 2) x + k (k + 8)$ $t : y = - 12 x + 18 k \land t : y = - (k + 2) x + k (k + 8) \Rightarrow$ $\Rightarrow - 12 = - (k + 2) \land 18 k = k (k + 8) \Rightarrow k = 10$ $par : y = - x^{2} + 8 x + 80$ $t : y = - 12 x + 180$ $shaded area = \underset{k}{\int^{3 k / 2}} t d x - \underset{k}{\int^{3 k / 2}} par d x =$ $= \underset{10}{\int^{15}} (x^{2} - 20 x + 100) d x = \frac{125}{3}$
	Commented by Tawa1 last updated on 10/Jun/19

	$God bless you sir$
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