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Question Number 201659 by LimPorly last updated on 10/Dec/23
Find the shortest distance between   point A(3,2) and curve y=(√x) (x>0).
$${Find}\:{the}\:{shortest}\:{distance}\:{between}\: \\ $$$${point}\:{A}\left(\mathrm{3},\mathrm{2}\right)\:{and}\:{curve}\:{y}=\sqrt{{x}}\:\left({x}>\mathrm{0}\right). \\ $$
Answered by mr W last updated on 10/Dec/23
say the distance is s.  say the point on the curve is (p^2 , p)  Φ=s^2 =(p^2 −3)^2 +(p−2)^2   (dΦ/dp)=4p(p^2 −3)+2(p−2)=0  p^3 −((5p)/2)−1=0  ⇒p=((√(30))/3) sin ((π/3)+(1/3)sin^(−1) ((3(√(30)))/(25)))  s_(min) =(√(13−((75)/9) sin^2  ((π/3)+(1/3)sin^(−1) ((3(√(30)))/(25)))−(√(30)) sin ((π/3)+(1/3)sin^(−1) ((3(√(30)))/(25))) ))       ≈0.257 552 568
$${say}\:{the}\:{distance}\:{is}\:{s}. \\ $$$${say}\:{the}\:{point}\:{on}\:{the}\:{curve}\:{is}\:\left({p}^{\mathrm{2}} ,\:{p}\right) \\ $$$$\Phi={s}^{\mathrm{2}} =\left({p}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} +\left({p}−\mathrm{2}\right)^{\mathrm{2}} \\ $$$$\frac{{d}\Phi}{{dp}}=\mathrm{4}{p}\left({p}^{\mathrm{2}} −\mathrm{3}\right)+\mathrm{2}\left({p}−\mathrm{2}\right)=\mathrm{0} \\ $$$${p}^{\mathrm{3}} −\frac{\mathrm{5}{p}}{\mathrm{2}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{p}=\frac{\sqrt{\mathrm{30}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{30}}}{\mathrm{25}}\right) \\ $$$${s}_{{min}} =\sqrt{\mathrm{13}−\frac{\mathrm{75}}{\mathrm{9}}\:\mathrm{sin}^{\mathrm{2}} \:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{30}}}{\mathrm{25}}\right)−\sqrt{\mathrm{30}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{30}}}{\mathrm{25}}\right)\:} \\ $$$$\:\:\:\:\:\approx\mathrm{0}.\mathrm{257}\:\mathrm{552}\:\mathrm{568} \\ $$
Commented by mr W last updated on 10/Dec/23
Commented by LimPorly last updated on 10/Dec/23
Thank you sir
$${Thank}\:{you}\:{sir} \\ $$
Answered by mr W last updated on 11/Dec/23
y=x^2   A(2,3)  P(p,p^2 )  tan θ=2p  2=p+r sin θ   ...(i)  3=p^2 −r cos θ  ...(ii)  r^2 =(2−p)^2 +(p^2 −3)^2   ((2−p)/(p^2 −3))=2p  p^3 −((5p)/2)−1=0  ⇒p=((√(30))/3) sin ((π/3)+(1/3)sin^(−1) ((3(√(30)))/(25)))
$${y}={x}^{\mathrm{2}} \\ $$$${A}\left(\mathrm{2},\mathrm{3}\right) \\ $$$${P}\left({p},{p}^{\mathrm{2}} \right) \\ $$$$\mathrm{tan}\:\theta=\mathrm{2}{p} \\ $$$$\mathrm{2}={p}+{r}\:\mathrm{sin}\:\theta\:\:\:…\left({i}\right) \\ $$$$\mathrm{3}={p}^{\mathrm{2}} −{r}\:\mathrm{cos}\:\theta\:\:…\left({ii}\right) \\ $$$${r}^{\mathrm{2}} =\left(\mathrm{2}−{p}\right)^{\mathrm{2}} +\left({p}^{\mathrm{2}} −\mathrm{3}\right)^{\mathrm{2}} \\ $$$$\frac{\mathrm{2}−{p}}{{p}^{\mathrm{2}} −\mathrm{3}}=\mathrm{2}{p} \\ $$$${p}^{\mathrm{3}} −\frac{\mathrm{5}{p}}{\mathrm{2}}−\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{p}=\frac{\sqrt{\mathrm{30}}}{\mathrm{3}}\:\mathrm{sin}\:\left(\frac{\pi}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{sin}^{−\mathrm{1}} \frac{\mathrm{3}\sqrt{\mathrm{30}}}{\mathrm{25}}\right) \\ $$

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