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Question Number 201890 by sonukgindia last updated on 15/Dec/23

	Answered by mr W last updated on 15/Dec/23

	$B^{'} = (- 2, 1)$ $\frac{y_{P} - 1}{0 - (- 2)} = \frac{5 - 1}{5 - (- 2)}$ $\Rightarrow y_{P} = \frac{15}{7} \Rightarrow P (0, \frac{15}{7})$ ${(d_{1} + d_{2})}_{m i n} = {A B}^{'} = \sqrt{7^{2} + 4^{2}} = \sqrt{65}$
	Commented by ajfour last updated on 15/Dec/23

	$W h a t i f i n s t e a d o f y a x i s (x = 0),$ $P l i e s o n p a r a b o l a x = \sqrt{y}; w h a t i s$ $d_{m i n} = {(d_{1} + d_{2})}_{m i n} = ?$
	Commented by ajfour last updated on 16/Dec/23
	$Say P(x, x^2 ) ; d_1 +d_2 =s(x) s(x)=(√((5−x)^2 +(5−x^2 )^2 ))+(√((x−2)^2 +(x^2 −1)^2 )) s_(min) =s(x_0 ) . Find x_0 . s^2 =h+k+2(√(hk)) h−k=24−8x^2 +21−6x h+k=2x^4 −10x^2 −14x+55 (ds/dx)=((((dh/dx)))/(2(√h)))+((((dk/dx)))/(2(√k)))=0 ⇒ (k/h)=−(dk/dh) ⇒ kdh+hdk=0 ⇒ (dh/h)+(dk/k)=0 ln (hk)=c=ln (h_0 k_0 )=ln 250 ⇒ hk=250 (k/h)=−{((2(x−2)+4x(x^2 −1))/(2(x−5)+4x(x^2 −5)))} s^2 −10(√(10))=h+k also (dh/dx)+(dk/dx)=0 ⇒ 8x^3 −20x−14=0 or (2x)^3 −10(2x)−14=0 x=(9^(2/3) /(18)){(63+(√(969)))^(1/3) +(63−(√(969)))^(1/3) } ≈ 1.855562$
	$S a y P (x, x^{2}); d_{1} + d_{2} = s (x)$ $s (x) = \sqrt{{(5 - x)}^{2} + {(5 - x^{2})}^{2}} + \sqrt{{(x - 2)}^{2} + {(x^{2} - 1)}^{2}}$ $s_{m i n} = s (x_{0}) . F i n d x_{0} .$ $s^{2} = h + k + 2 \sqrt{h k}$ $h - k = 24 - 8 x^{2} + 21 - 6 x$ $h + k = 2 x^{4} - 10 x^{2} - 14 x + 55$ $\frac{d s}{d x} = \frac{(\frac{d h}{d x})}{2 \sqrt{h}} + \frac{(\frac{d k}{d x})}{2 \sqrt{k}} = 0$ $\Rightarrow \frac{k}{h} = - \frac{d k}{d h}$ $\Rightarrow k d h + h d k = 0$ $\Rightarrow \frac{d h}{h} + \frac{d k}{k} = 0$ $\ln (h k) = c = \ln (h_{0} k_{0}) = \ln 250$ $\Rightarrow h k = 250$ $\frac{k}{h} = - {\frac{2 (x - 2) + 4 x (x^{2} - 1)}{2 (x - 5) + 4 x (x^{2} - 5)}}$ $s^{2} - 10 \sqrt{10} = h + k$ $a l s o \frac{d h}{d x} + \frac{d k}{d x} = 0$ $\Rightarrow 8 x^{3} - 20 x - 14 = 0$ $o r {(2 x)}^{3} - 10 (2 x) - 14 = 0$ $x = \frac{9^{2 / 3}}{18} {{(63 + \sqrt{969})}^{1 / 3} + {(63 - \sqrt{969})}^{1 / 3}}$ $\approx 1.855562$
	Commented by mr W last updated on 15/Dec/23

	$P (p, p^{2})$ $t a n g e n t t h r o u g h P :$ $y = p^{2} + 2 p (x - p)$ $\Rightarrow 2 p x - y^{2} - p^{2} = 0$ $i m a g e o f B (2, 1) i s B^{'} (h, k)$ $\frac{h - 2}{2 p} = \frac{k - 1}{- 1} = \frac{- 2 (2 p \times 2 - 1^{2} - p^{2})}{{(2 p)}^{2} + {(- 1)}^{2}}$ $\Rightarrow h = 2 + \frac{4 p (p^{2} - 4 p + 1)}{4 p^{2} + 1}$ $\Rightarrow k = 1 - \frac{2 (p^{2} - 4 p + 1)}{4 p^{2} + 1}$ $\frac{p - h}{5 - h} = \frac{p^{2} - k}{5 - k}$ $\frac{p - 5}{- 4 p^{3} + 28 p^{2} - 4 p + 3} = \frac{p^{2} - 5}{18 p^{2} - 8 p + 6}$ $4 p^{5} - 28 p^{4} + 2 p^{3} + 39 p^{2} + 26 p - 15 = 0$ $\Rightarrow p \approx 1.55436$
	Commented by mr W last updated on 15/Dec/23

	Commented by ajfour last updated on 15/Dec/23

	$s p l e n d i d w a y! T h a n k s s i r .$
	Answered by cortano12 last updated on 15/Dec/23
	$let P(0,b) ⇒AP=(√(5^2 +(5−b)^2 )) ⇒BP=(√(2^2 +(1−b)^2 )) ⇒f(b)=(√(25+(5−b)^2 ))+(√(4+(1−b)^2 )) ⇒ f ′(b)= ((−(5−b))/( (√(25+(5−b)^2 )))) −(((1−b))/( (√(4+(1−b)^2 )))) ⇒f ′(b)= 0 ((b−5)/( (√(25+(5−b)^2 )))) = ((1−b)/( (√(4+(1−b)^2 )))) ((b^2 −10b+25)/( 50−10b+b^2 )) = ((b^2 −2b+1)/(5−2b+b^2 )) { ((b_1 =−(5/3))),((b_2 =((15)/7))) :} { ((P_1 (0,−(5/3)))),((P_2 (0,((15)/7)))) :}⇒ { ((d_1 =(√(25+((400)/9)))+(√(4+((64)/9)))=((35)/3)≈11.67)),((d_2 =(√(25+((400)/(49))))+(√(4+((64)/(49))))=(√(65))≈8.062)) :} ∴ d_(min) = (√(65)) at P(0,((15)/7))$
	$let P (0, b)$ $\Rightarrow AP = \sqrt{5^{2} + {(5 - b)}^{2}}$ $\Rightarrow BP = \sqrt{2^{2} + {(1 - b)}^{2}}$ $\Rightarrow f (b) = \sqrt{25 + {(5 - b)}^{2}} + \sqrt{4 + {(1 - b)}^{2}}$ $\Rightarrow f^{'} (b) = \frac{- (5 - b)}{\sqrt{25 + {(5 - b)}^{2}}} - \frac{(1 - b)}{\sqrt{4 + {(1 - b)}^{2}}}$ $\Rightarrow f^{'} (b) = 0$ $\frac{b - 5}{\sqrt{25 + {(5 - b)}^{2}}} = \frac{1 - b}{\sqrt{4 + {(1 - b)}^{2}}}$ $\frac{b^{2} - 10 b + 25}{50 - 10 b + b^{2}} = \frac{b^{2} - 2 b + 1}{5 - 2 b + b^{2}}$ ${\begin{cases} b_{1} = - \frac{5}{3} \\ b_{2} = \frac{15}{7} \end{cases}$ ${\begin{cases} P_{1} (0, - \frac{5}{3}) \\ P_{2} (0, \frac{15}{7}) \end{cases} \Rightarrow {\begin{cases} d_{1} = \sqrt{25 + \frac{400}{9}} + \sqrt{4 + \frac{64}{9}} = \frac{35}{3} \approx 11.67 \\ d_{2} = \sqrt{25 + \frac{400}{49}} + \sqrt{4 + \frac{64}{49}} = \sqrt{65} \approx 8.062 \end{cases}$ $∴ d_{\min} = \sqrt{65} at P (0, \frac{15}{7})$
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