Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

Question Number 56914 by rahul 19 last updated on 26/Mar/19

The shortest distance between the point ((3/2),0) and the curve y=(√x) ,(x>0) is ?

$${The}\:{shortest}\:{distance}\:{between}\:{the}\:{point} \\ $$ $$\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{0}\right)\:{and}\:{the}\:{curve}\:{y}=\sqrt{{x}}\:,\left({x}>\mathrm{0}\right)\:{is}\:? \\ $$

Answered by Smail last updated on 26/Mar/19

d=(√((x_2 −x_1 )^2 +(y_2 −y_1 )^2 )) =(√((x−(3/2))^2 +((√x)−0)^2 ))=(√(x^2 −2x+(9/4))) d′=((2x−2)/(2(√(x^2 −2x+(9/2))))) CV is x=1 d_(min) =(√(1^2 −2×1+(9/4)))=((√5)/2)

$${d}=\sqrt{\left({x}_{\mathrm{2}} −{x}_{\mathrm{1}} \right)^{\mathrm{2}} +\left({y}_{\mathrm{2}} −{y}_{\mathrm{1}} \right)^{\mathrm{2}} } \\ $$ $$=\sqrt{\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{{x}}−\mathrm{0}\right)^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{4}}} \\ $$ $${d}'=\frac{\mathrm{2}{x}−\mathrm{2}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{2}}}} \\ $$ $${CV}\:{is}\:\:{x}=\mathrm{1}\:\: \\ $$ $${d}_{{min}} =\sqrt{\mathrm{1}^{\mathrm{2}} −\mathrm{2}×\mathrm{1}+\frac{\mathrm{9}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Commented byrahul 19 last updated on 26/Mar/19

thank you sir!

Answered by tanmay.chaudhury50@gmail.com last updated on 26/Mar/19

let (x,(√x) )be a point on the curve. distance l=(√((x−(3/2))^2 +((√x) −0)^2 )) l^2 =x^2 −2×x×(3/2)+(9/4)+x l^2 =x^2 −2x+(9/4) l=(√(x^2 −2x+(9/4))) (dl/dx)=(1/(2(√(x^2 −2x+(9/4)))))×(2x−2) (dl/dx)=((x−1)/(√(x^2 −2x+(9/4)))) for max/min (dl/dx)=0 so x=1 (d^2 l/dx^2 )=(((√(x^2 −2x+(9/4) )) ×1−(x−1)×(1/(2(√(x^2 −2x+(9/4) ))))×(2x−2))/((x^2 −2x+(9/4)))) ((d^2 l/dx^2 ))_(x=1) =(((√(5/4)) ×1)/(((5/4))))=(√(4/5)) >0 so minimum distance is between point(x,(√x) ) that means (1,1) and((3/2),0) is l=(√(((3/2)−1)^2 +(1−0)^2 )) =(√((1/4)+1)) =(√(5/4)) =((√5)/2)←this is the snswer

$${let}\:\left({x},\sqrt{{x}}\:\right){be}\:{a}\:{point}\:{on}\:{the}\:{curve}. \\ $$ $${distance}\:{l}=\sqrt{\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{{x}}\:−\mathrm{0}\right)^{\mathrm{2}} }\: \\ $$ $${l}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{2}×{x}×\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{9}}{\mathrm{4}}+{x} \\ $$ $${l}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{4}} \\ $$ $${l}=\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{4}}}\: \\ $$ $$\frac{{dl}}{{dx}}=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{4}}}}×\left(\mathrm{2}{x}−\mathrm{2}\right) \\ $$ $$\frac{{dl}}{{dx}}=\frac{{x}−\mathrm{1}}{\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{4}}}} \\ $$ $${for}\:{max}/{min}\:\frac{{dl}}{{dx}}=\mathrm{0}\:\:\:{so}\:{x}=\mathrm{1} \\ $$ $$ \\ $$ $$\frac{{d}^{\mathrm{2}} {l}}{{dx}^{\mathrm{2}} }=\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{4}}\:}\:×\mathrm{1}−\left({x}−\mathrm{1}\right)×\frac{\mathrm{1}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{4}}\:}}×\left(\mathrm{2}{x}−\mathrm{2}\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{4}}\right)} \\ $$ $$\left(\frac{{d}^{\mathrm{2}} {l}}{{dx}^{\mathrm{2}} }\right)_{{x}=\mathrm{1}} =\frac{\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}\:×\mathrm{1}}{\left(\frac{\mathrm{5}}{\mathrm{4}}\right)}=\sqrt{\frac{\mathrm{4}}{\mathrm{5}}}\:>\mathrm{0} \\ $$ $${so}\:{minimum}\:{distance}\:{is}\:{between}\:{point}\left({x},\sqrt{{x}}\:\right) \\ $$ $${that}\:{means}\:\left(\mathrm{1},\mathrm{1}\right)\:{and}\left(\frac{\mathrm{3}}{\mathrm{2}},\mathrm{0}\right)\:{is} \\ $$ $${l}=\sqrt{\left(\frac{\mathrm{3}}{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{1}−\mathrm{0}\right)^{\mathrm{2}} }\: \\ $$ $$=\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\mathrm{1}}\: \\ $$ $$=\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}\: \\ $$ $$=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\leftarrow{this}\:{is}\:{the}\:{snswer} \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$

Commented byrahul 19 last updated on 26/Mar/19

thank you sir!

Answered by mr W last updated on 26/Mar/19

d=(√((x−(3/2))^2 +((√x)−0)^2 ))=(√(x^2 −2x+(9/4))) =(√((x−1)^2 +(5/4)))≥(√(5/4))=((√5)/2) ⇒d_(min) =((√5)/2) at x=1

$${d}=\sqrt{\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\sqrt{{x}}−\mathrm{0}\right)^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{x}+\frac{\mathrm{9}}{\mathrm{4}}} \\ $$ $$=\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\frac{\mathrm{5}}{\mathrm{4}}}\geqslant\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}=\frac{\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$ $$\Rightarrow{d}_{{min}} =\frac{\sqrt{\mathrm{5}}}{\mathrm{2}}\:{at}\:{x}=\mathrm{1} \\ $$

Commented byrahul 19 last updated on 26/Mar/19

thank you sir!

Terms of Service

Privacy Policy

Contact: info@tinkutara.com