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Question Number 205594 by cortano12 last updated on 25/Mar/24

	Answered by Red1ight last updated on 25/Mar/24

	$y^{2} = 4 x$ $y = 2 \sqrt{x}$ $\frac{d y}{d x} = \frac{1}{\sqrt{x}} = \frac{2}{y}$ $(x_{0}, y_{0}) = (2, 8), m = - \sqrt{x}$ $y = 8 + 2 \sqrt{x} - x \sqrt{x}$ $8 + 2 \sqrt{x} - x \sqrt{x} = 2 \sqrt{x}$ $8 - x \sqrt{x} = 0$ $x^{\frac{3}{2}} = 8$ $x = 4$ $C = (4, 4)$ $B = (x_{b}, y_{b}), m = - 2$ $x_{b} = 2 + \cos (\tan^{- 1} (2)) = 2 + \frac{\sqrt{5}}{5}$ $y_{b} = 8 + \sin (\tan^{- 1} (2)) = 8 + \frac{- 2 \sqrt{5}}{5}$ $B \approx (2.447, 7.105)$ $B C \approx 7.067$
	Commented by Red1ight last updated on 25/Mar/24

	Commented by Red1ight last updated on 25/Mar/24

	$Switched B and C$
	Answered by mr W last updated on 25/Mar/24

	Commented by mr W last updated on 25/Mar/24

	$A (2, 8)$ $P (p, q)$ $4 p = q^{2}$ $Φ = R^{2} = {(p - 2)}^{2} + {(q - 8)}^{2} = {(\frac{q^{2}}{4} - 2)}^{2} + {(q - 8)}^{2}$ $\frac{d Φ}{d q} = q (\frac{q^{2}}{4} - 2) + 2 (q - 8) = 0$ $\Rightarrow q^{3} - 64 = 0$ $\Rightarrow q = 4 \Rightarrow p = 4$ $\Rightarrow R_{m i n}^{2} = {(4 - 2)}^{2} + {(4 - 8)}^{2} = 20 \Rightarrow R_{m i n} = 2 \sqrt{5}$ $m i n B C = R_{m i n} - 1 = 2 \sqrt{5} - 1 \approx 3.472 ✓$
	Commented by mr W last updated on 25/Mar/24

	Answered by mr W last updated on 25/Mar/24
	$Method II B(2+cos θ, 8+sin θ) C((y^2 /4), y) Φ=BC^2 =(2+cos θ−(y^2 /4))^2 +(8+sin θ−y)^2 (∂Φ/∂θ)=−2sin θ(2+cos θ−(y^2 /4))+2cos θ(8+sin θ−y)=0 sin θ(2+cos θ−(y^2 /4))=cos θ(8+sin θ−y)=0 ⇒tan θ=((4(8−y))/(8−y^2 )) (∂Φ/∂y)=−y(2+cos θ−(y^2 /4))−2(8+sin θ−y)=0 −y(2+cos θ−(y^2 /4))=2(8+sin θ−y) ⇒−(y/2)=((sin θ)/(cos θ))=tan θ ((4(8−y))/(8−y^2 ))=−(y/2) y^3 =64 ⇒y=4 ⇒tan θ=−2 ⇒ { ((cos θ=−(1/( (√5))) or (1/( (√5))))),((sin θ=(2/( (√5))) or −(2/( (√5))))) :} BC_(min) ^2 =(2+(1/( (√5)))−(4^2 /4))^2 +(8−(2/( (√5)))−4)^2 =21−4(√5) ⇒(BC)_(min) =(√(21−4(√5)))=2(√5)−1 ✓$
	$M e t h o d I I$ $B (2 + \cos θ, 8 + \sin θ)$ $C (\frac{y^{2}}{4}, y)$ $Φ = {B C}^{2} = {(2 + \cos θ - \frac{y^{2}}{4})}^{2} + {(8 + \sin θ - y)}^{2}$ $\frac{\partial Φ}{\partial θ} = - 2 \sin θ (2 + \cos θ - \frac{y^{2}}{4}) + 2 \cos θ (8 + \sin θ - y) = 0$ $\sin θ (2 + \cos θ - \frac{y^{2}}{4}) = \cos θ (8 + \sin θ - y) = 0$ $\Rightarrow \tan θ = \frac{4 (8 - y)}{8 - y^{2}}$ $\frac{\partial Φ}{\partial y} = - y (2 + \cos θ - \frac{y^{2}}{4}) - 2 (8 + \sin θ - y) = 0$ $- y (2 + \cos θ - \frac{y^{2}}{4}) = 2 (8 + \sin θ - y)$ $\Rightarrow - \frac{y}{2} = \frac{\sin θ}{\cos θ} = \tan θ$ $\frac{4 (8 - y)}{8 - y^{2}} = - \frac{y}{2}$ $y^{3} = 64 \Rightarrow y = 4$ $\Rightarrow \tan θ = - 2 \Rightarrow {\begin{cases} \cos θ = - \frac{1}{\sqrt{5}} o r \frac{1}{\sqrt{5}} \\ \sin θ = \frac{2}{\sqrt{5}} o r - \frac{2}{\sqrt{5}} \end{cases}$ ${B C}_{m i n}^{2} = {(2 + \frac{1}{\sqrt{5}} - \frac{4^{2}}{4})}^{2} + {(8 - \frac{2}{\sqrt{5}} - 4)}^{2} = 21 - 4 \sqrt{5}$ $\Rightarrow {(B C)}_{m i n} = \sqrt{21 - 4 \sqrt{5}} = 2 \sqrt{5} - 1 ✓$
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