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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) ((y_P −1)/(0−(−2)))=((5−1)/(5−(−2))) ⇒y_P =((15)/7) ⇒P(0, ((15)/7)) (d_1 +d_2 )_(min) =AB′=(√(7^2 +4^2 ))=(√(65))
$${B}'=\left(−\mathrm{2},\mathrm{1}\right) \\ $$$$\frac{{y}_{{P}} −\mathrm{1}}{\mathrm{0}−\left(−\mathrm{2}\right)}=\frac{\mathrm{5}−\mathrm{1}}{\mathrm{5}−\left(−\mathrm{2}\right)} \\ $$$$\Rightarrow{y}_{{P}} =\frac{\mathrm{15}}{\mathrm{7}}\:\Rightarrow{P}\left(\mathrm{0},\:\frac{\mathrm{15}}{\mathrm{7}}\right) \\ $$$$\left({d}_{\mathrm{1}} +{d}_{\mathrm{2}} \right)_{{min}} ={AB}'=\sqrt{\mathrm{7}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\sqrt{\mathrm{65}} \\ $$
Commented by ajfour last updated on 15/Dec/23
What if instead of y axis (x=0), P lies on parabola x=(√y) ; what is d_(min) =(d_1 +d_2 )_(min) =?
$${What}\:{if}\:{instead}\:{of}\:{y}\:{axis}\:\left({x}=\mathrm{0}\right), \\ $$$${P}\:{lies}\:{on}\:\:{parabola}\:\:{x}=\sqrt{{y}}\:;\:{what}\:{is} \\ $$$${d}_{{min}} =\left({d}_{\mathrm{1}} +{d}_{\mathrm{2}} \right)_{{min}} =?\:\: \\ $$
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
$${Say}\:{P}\left({x},\:{x}^{\mathrm{2}} \right)\:\:\:\:;\:\:\:\:\:{d}_{\mathrm{1}} +{d}_{\mathrm{2}} ={s}\left({x}\right) \\ $$$${s}\left({x}\right)=\sqrt{\left(\mathrm{5}−{x}\right)^{\mathrm{2}} +\left(\mathrm{5}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }+\sqrt{\left({x}−\mathrm{2}\right)^{\mathrm{2}} +\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${s}_{{min}} ={s}\left({x}_{\mathrm{0}} \right)\:\:.\:{Find}\:{x}_{\mathrm{0}} . \\ $$$${s}^{\mathrm{2}} ={h}+{k}+\mathrm{2}\sqrt{{hk}} \\ $$$${h}−{k}=\mathrm{24}−\mathrm{8}{x}^{\mathrm{2}} +\mathrm{21}−\mathrm{6}{x} \\ $$$${h}+{k}=\mathrm{2}{x}^{\mathrm{4}} −\mathrm{10}{x}^{\mathrm{2}} −\mathrm{14}{x}+\mathrm{55} \\ $$$$\frac{{ds}}{{dx}}=\frac{\left(\frac{{dh}}{{dx}}\right)}{\mathrm{2}\sqrt{{h}}}+\frac{\left(\frac{{dk}}{{dx}}\right)}{\mathrm{2}\sqrt{{k}}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\frac{{k}}{{h}}=−\frac{{dk}}{{dh}} \\ $$$$\Rightarrow\:\:{kdh}+{hdk}=\mathrm{0} \\ $$$$\Rightarrow\:\:\frac{{dh}}{{h}}+\frac{{dk}}{{k}}=\mathrm{0} \\ $$$$\mathrm{ln}\:\left({hk}\right)={c}=\mathrm{ln}\:\left({h}_{\mathrm{0}} {k}_{\mathrm{0}} \right)=\mathrm{ln}\:\mathrm{250} \\ $$$$\Rightarrow\:\:{hk}=\mathrm{250} \\ $$$$\:\:\:\:\:\:\frac{{k}}{{h}}=−\left\{\frac{\mathrm{2}\left({x}−\mathrm{2}\right)+\mathrm{4}{x}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}\left({x}−\mathrm{5}\right)+\mathrm{4}{x}\left({x}^{\mathrm{2}} −\mathrm{5}\right)}\right\} \\ $$$${s}^{\mathrm{2}} −\mathrm{10}\sqrt{\mathrm{10}}={h}+{k} \\ $$$${also}\:\:\:\frac{{dh}}{{dx}}+\frac{{dk}}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow\:\:\mathrm{8}{x}^{\mathrm{3}} −\mathrm{20}{x}−\mathrm{14}=\mathrm{0} \\ $$$${or}\:\:\left(\mathrm{2}{x}\right)^{\mathrm{3}} −\mathrm{10}\left(\mathrm{2}{x}\right)−\mathrm{14}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{9}^{\mathrm{2}/\mathrm{3}} }{\mathrm{18}}\left\{\left(\mathrm{63}+\sqrt{\mathrm{969}}\right)^{\mathrm{1}/\mathrm{3}} +\left(\mathrm{63}−\sqrt{\mathrm{969}}\right)^{\mathrm{1}/\mathrm{3}} \right\} \\ $$$$\:\approx\:\mathrm{1}.\mathrm{855562} \\ $$
Commented by mr W last updated on 15/Dec/23
P(p, p^2 ) tangent through P: y=p^2 +2p(x−p) ⇒2px−y^2 −p^2 =0 image of B(2, 1) is B′(h, k) ((h−2)/(2p))=((k−1)/(−1))=((−2(2p×2−1^2 −p^2 ))/((2p)^2 +(−1)^2 )) ⇒h=2+((4p(p^2 −4p+1))/(4p^2 +1)) ⇒k=1−((2(p^2 −4p+1))/(4p^2 +1)) ((p−h)/(5−h))=((p^2 −k)/(5−k)) ((p−5)/(−4p^3 +28p^2 −4p+3))=((p^2 −5)/(18p^2 −8p+6)) 4p^5 −28p^4 +2p^3 +39p^2 +26p−15=0 ⇒p≈1.55436
$${P}\left({p},\:{p}^{\mathrm{2}} \right) \\ $$$${tangent}\:{through}\:{P}: \\ $$$${y}={p}^{\mathrm{2}} +\mathrm{2}{p}\left({x}−{p}\right) \\ $$$$\Rightarrow\mathrm{2}{px}−{y}^{\mathrm{2}} −{p}^{\mathrm{2}} =\mathrm{0} \\ $$$${image}\:{of}\:{B}\left(\mathrm{2},\:\mathrm{1}\right)\:{is}\:{B}'\left({h},\:{k}\right) \\ $$$$\frac{{h}−\mathrm{2}}{\mathrm{2}{p}}=\frac{{k}−\mathrm{1}}{−\mathrm{1}}=\frac{−\mathrm{2}\left(\mathrm{2}{p}×\mathrm{2}−\mathrm{1}^{\mathrm{2}} −{p}^{\mathrm{2}} \right)}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} +\left(−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{h}=\mathrm{2}+\frac{\mathrm{4}{p}\left({p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{1}\right)}{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\Rightarrow{k}=\mathrm{1}−\frac{\mathrm{2}\left({p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{1}\right)}{\mathrm{4}{p}^{\mathrm{2}} +\mathrm{1}} \\ $$$$\frac{{p}−{h}}{\mathrm{5}−{h}}=\frac{{p}^{\mathrm{2}} −{k}}{\mathrm{5}−{k}} \\ $$$$\frac{{p}−\mathrm{5}}{−\mathrm{4}{p}^{\mathrm{3}} +\mathrm{28}{p}^{\mathrm{2}} −\mathrm{4}{p}+\mathrm{3}}=\frac{{p}^{\mathrm{2}} −\mathrm{5}}{\mathrm{18}{p}^{\mathrm{2}} −\mathrm{8}{p}+\mathrm{6}} \\ $$$$\mathrm{4}{p}^{\mathrm{5}} −\mathrm{28}{p}^{\mathrm{4}} +\mathrm{2}{p}^{\mathrm{3}} +\mathrm{39}{p}^{\mathrm{2}} +\mathrm{26}{p}−\mathrm{15}=\mathrm{0} \\ $$$$\Rightarrow{p}\approx\mathrm{1}.\mathrm{55436} \\ $$
Commented by mr W last updated on 15/Dec/23
Commented by ajfour last updated on 15/Dec/23
splendid way! Thanks sir.
$${splendid}\:{way}!\:{Thanks}\:{sir}. \\ $$
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))
$$\:\mathrm{let}\:\mathrm{P}\left(\mathrm{0},\mathrm{b}\right) \\ $$$$\:\Rightarrow\mathrm{AP}=\sqrt{\mathrm{5}^{\mathrm{2}} +\left(\mathrm{5}−\mathrm{b}\right)^{\mathrm{2}} } \\ $$$$\:\Rightarrow\mathrm{BP}=\sqrt{\mathrm{2}^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{b}\right)^{\mathrm{2}} } \\ $$$$\:\Rightarrow\mathrm{f}\left(\mathrm{b}\right)=\sqrt{\mathrm{25}+\left(\mathrm{5}−\mathrm{b}\right)^{\mathrm{2}} }+\sqrt{\mathrm{4}+\left(\mathrm{1}−\mathrm{b}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow\:\mathrm{f}\:'\left(\mathrm{b}\right)=\:\frac{−\left(\mathrm{5}−\mathrm{b}\right)}{\:\sqrt{\mathrm{25}+\left(\mathrm{5}−\mathrm{b}\right)^{\mathrm{2}} }}\:−\frac{\left(\mathrm{1}−\mathrm{b}\right)}{\:\sqrt{\mathrm{4}+\left(\mathrm{1}−\mathrm{b}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\mathrm{f}\:'\left(\mathrm{b}\right)=\:\mathrm{0}\: \\ $$$$\:\frac{\mathrm{b}−\mathrm{5}}{\:\sqrt{\mathrm{25}+\left(\mathrm{5}−\mathrm{b}\right)^{\mathrm{2}} }}\:=\:\frac{\mathrm{1}−\mathrm{b}}{\:\sqrt{\mathrm{4}+\left(\mathrm{1}−\mathrm{b}\right)^{\mathrm{2}} }} \\ $$$$\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{10b}+\mathrm{25}}{\:\mathrm{50}−\mathrm{10b}+\mathrm{b}^{\mathrm{2}} }\:=\:\frac{\mathrm{b}^{\mathrm{2}} −\mathrm{2b}+\mathrm{1}}{\mathrm{5}−\mathrm{2b}+\mathrm{b}^{\mathrm{2}} } \\ $$$$\:\begin{cases}{\boldsymbol{\mathrm{b}}_{\mathrm{1}} =−\frac{\mathrm{5}}{\mathrm{3}}}\\{\mathrm{b}_{\mathrm{2}} =\frac{\mathrm{15}}{\mathrm{7}}}\end{cases} \\ $$$$\:\begin{cases}{\boldsymbol{\mathrm{P}}_{\mathrm{1}} \left(\mathrm{0},−\frac{\mathrm{5}}{\mathrm{3}}\right)}\\{\mathrm{P}_{\mathrm{2}} \left(\mathrm{0},\frac{\mathrm{15}}{\mathrm{7}}\right)}\end{cases}\Rightarrow\begin{cases}{\mathrm{d}_{\mathrm{1}} =\sqrt{\mathrm{25}+\frac{\mathrm{400}}{\mathrm{9}}}+\sqrt{\mathrm{4}+\frac{\mathrm{64}}{\mathrm{9}}}=\frac{\mathrm{35}}{\mathrm{3}}\approx\mathrm{11}.\mathrm{67}}\\{\mathrm{d}_{\mathrm{2}} =\sqrt{\mathrm{25}+\frac{\mathrm{400}}{\mathrm{49}}}+\sqrt{\mathrm{4}+\frac{\mathrm{64}}{\mathrm{49}}}=\sqrt{\mathrm{65}}\approx\mathrm{8}.\mathrm{062}}\end{cases} \\ $$$$\:\therefore\:\mathrm{d}_{\mathrm{min}} =\:\sqrt{\mathrm{65}}\:\mathrm{at}\:\mathrm{P}\left(\mathrm{0},\frac{\mathrm{15}}{\mathrm{7}}\right) \\ $$

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