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Question Number 135004 by bramlexs22 last updated on 09/Mar/21

$$ \\ $$ the closest distance from the point on the curve y = x ^ 3-1 to the curve x = y ^2+ 4 is equal to\\n

Commented bybramlexs22 last updated on 09/Mar/21

Commented byEDWIN88 last updated on 09/Mar/21

step(1) slope of tangent line curve y=x^3 −1 at point P(a,a^3 −1) is m_1 =3a^2 then slope of normal line is −(1/(3a^2 )) the equation of normal line at point (a,a^3 −1) is y=−(1/(3a^2 ))(x−a)+a^3 −1⇒y =−(x/(3a^2 ))+(1/(3a))+a^3 −1 step(2) slope of tangent line curve y^2 =x−4 at point Q(b^2 +4, b) is m_2 = (1/(2b)) , then slope of normal line is −2b , and the equation of normal line equal to y = −2b(x−b^2 −4)+b⇒y=−2bx+2b^3 +9b step(3) the both of equation of normal line is same , we get { ((−(1/(3a^2 )) = −2b ; a^2 b = (1/6))),(((1/(3a))+a^3 −1 = 2b^3 +9b)) :} solving for a and b , ⇒(1/(3a))+a^3 −1=2((1/(6a^2 )))^3 +9((1/(6a^2 ))) ⇒(1/(3a))+a^3 −1 = (1/(108a^6 )) +(3/(2a^2 )) 36a^5 +108a^9 −108a^6 =1+162a^4 108a^9 −108a^6 +36a^5 −162a^4 −1=0 we get { ((a=1.2067)),((b = 0.1145)) :} we get a point P(1.2067; 0.7571) and Q(4.0131; 0.1145) therefore the closest distance is ∣PQ∣ = (√((4.0131−1.2067)^2 +(0.1145−0.7571)^2 )) ∣PQ∣ ≈ 2.87903

$$\mathrm{step}\left(\mathrm{1}\right)\:\mathrm{slope}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{3}} −\mathrm{1}\:\mathrm{at}\:\mathrm{point}\:\mathrm{\color{mathbrown}{P}}\left(\mathrm{a},\mathrm{a}^{\mathrm{3}} −\mathrm{1}\right) \\ $$ $$\mathrm{is}\:\mathrm{m}_{\mathrm{\color{mathbrown}{1}}} =\mathrm{3a}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\:\mathrm{is}\:−\frac{\mathrm{1}}{\mathrm{3a}^{\mathrm{2}} } \\ $$ $$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\:\mathrm{at}\:\mathrm{point}\:\left(\mathrm{a},\mathrm{a}^{\mathrm{3}} −\mathrm{1}\right) \\ $$ $$\mathrm{is}\:\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{3a}^{\mathrm{2}} }\left(\mathrm{x}−\mathrm{a}\right)+\mathrm{a}^{\mathrm{3}} −\mathrm{1}\color{mathred}{\Rightarrow}\mathrm{\color{mathred}{y}}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{−}\frac{\mathrm{\color{mathred}{x}}}{\mathrm{\color{mathred}{3}\color{mathred}{a}}^{\mathrm{\color{mathred}{2}}} }\color{mathred}{+}\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{3}\color{mathred}{a}}}\color{mathred}{+}\mathrm{\color{mathred}{a}}^{\mathrm{\color{mathred}{3}}} \color{mathred}{−}\mathrm{\color{mathred}{1}} \\ $$ $$\mathrm{step}\left(\mathrm{2}\right)\:\mathrm{slope}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{curve}\:\mathrm{y}^{\mathrm{2}} =\mathrm{x}−\mathrm{4}\:\mathrm{at}\:\mathrm{point} \\ $$ $$\mathrm{\color{mathbrown}{Q}}\left(\mathrm{b}^{\mathrm{2}} +\mathrm{4},\:\mathrm{b}\right)\:\mathrm{is}\:\mathrm{\color{mathbrown}{m}}_{\mathrm{\color{mathbrown}{2}}} \:=\:\frac{\mathrm{\color{mathbrown}{1}}}{\mathrm{\color{mathbrown}{2}\color{mathbrown}{b}}}\color{mathbrown}{\:},\:\mathrm{then}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{normal} \\ $$ $$\mathrm{line}\:\mathrm{is}\:−\mathrm{2b}\:,\:\mathrm{and}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line} \\ $$ $$\mathrm{equal}\:\mathrm{to}\:\mathrm{\color{mathred}{y}}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:}\color{mathred}{−}\mathrm{\color{mathred}{2}\color{mathred}{b}}\color{mathred}{\left(}\mathrm{\color{mathred}{x}}\color{mathred}{−}\mathrm{\color{mathred}{b}}^{\mathrm{\color{mathred}{2}}} \color{mathred}{−}\mathrm{\color{mathred}{4}}\color{mathred}{\right)}\color{mathred}{+}\mathrm{\color{mathred}{b}}\color{mathred}{\Rightarrow}\mathrm{\color{mathred}{y}}\color{mathred}{=}\color{mathred}{−}\mathrm{\color{mathred}{2}\color{mathred}{b}\color{mathred}{x}}\color{mathred}{+}\mathrm{\color{mathred}{2}\color{mathred}{b}}^{\mathrm{\color{mathred}{3}}} \color{mathred}{+}\mathrm{\color{mathred}{9}\color{mathred}{b}} \\ $$ $$\mathrm{step}\left(\mathrm{3}\right)\:\mathrm{the}\:\mathrm{both}\:\mathrm{of}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line} \\ $$ $$\mathrm{is}\:\mathrm{same}\:\color{mathred}{,}\color{mathred}{\:}\mathrm{we}\:\mathrm{get}\:\begin{cases}{−\frac{\mathrm{1}}{\mathrm{3a}^{\mathrm{2}} }\:=\:−\mathrm{2b}\:;\:\mathrm{a}^{\mathrm{2}} \mathrm{b}\:=\:\frac{\mathrm{1}}{\mathrm{6}}}\\{\frac{\mathrm{1}}{\mathrm{3a}}+\mathrm{a}^{\mathrm{3}} −\mathrm{1}\:=\:\mathrm{2b}^{\mathrm{3}} +\mathrm{9b}}\end{cases} \\ $$ $$\mathrm{solving}\:\mathrm{for}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:,\:\Rightarrow\frac{\mathrm{1}}{\mathrm{3a}}+\mathrm{a}^{\mathrm{3}} −\mathrm{1}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{6a}^{\mathrm{2}} }\right)^{\mathrm{3}} +\mathrm{9}\left(\frac{\mathrm{1}}{\mathrm{6a}^{\mathrm{2}} }\right) \\ $$ $$\color{mathbrown}{\Rightarrow}\frac{\mathrm{1}}{\mathrm{3a}}+\mathrm{a}^{\mathrm{3}} −\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{108a}^{\mathrm{6}} }\:+\frac{\mathrm{3}}{\mathrm{2a}^{\mathrm{2}} } \\ $$ $$\mathrm{36a}^{\mathrm{5}} +\mathrm{108a}^{\mathrm{9}} −\mathrm{108a}^{\mathrm{6}} =\mathrm{1}+\mathrm{162a}^{\mathrm{4}} \\ $$ $$\mathrm{108a}^{\mathrm{9}} −\mathrm{108a}^{\mathrm{6}} +\mathrm{36a}^{\mathrm{5}} −\mathrm{162a}^{\mathrm{4}} −\mathrm{1}=\mathrm{0} \\ $$ $$\mathrm{we}\:\mathrm{get}\:\begin{cases}{\mathrm{a}=\mathrm{1}.\mathrm{2067}}\\{\mathrm{b}\:=\:\mathrm{0}.\mathrm{1145}}\end{cases} \\ $$ $$\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{point}\:\mathrm{P}\left(\mathrm{1}.\mathrm{2067};\:\mathrm{0}.\mathrm{7571}\right)\:\mathrm{and}\: \\ $$ $$\mathrm{Q}\left(\mathrm{4}.\mathrm{0131};\:\mathrm{0}.\mathrm{1145}\right) \\ $$ $$\mathrm{\color{mathbrown}{t}\color{mathbrown}{h}\color{mathbrown}{e}\color{mathbrown}{r}\color{mathbrown}{e}\color{mathbrown}{f}\color{mathbrown}{o}\color{mathbrown}{r}\color{mathbrown}{e}}\color{mathbrown}{\:}\mathrm{\color{mathbrown}{t}\color{mathbrown}{h}\color{mathbrown}{e}}\color{mathbrown}{\:}\mathrm{\color{mathbrown}{c}\color{mathbrown}{l}\color{mathbrown}{o}\color{mathbrown}{s}\color{mathbrown}{e}\color{mathbrown}{s}\color{mathbrown}{t}}\color{mathbrown}{\:}\mathrm{\color{mathbrown}{d}\color{mathbrown}{i}\color{mathbrown}{s}\color{mathbrown}{t}\color{mathbrown}{a}\color{mathbrown}{n}\color{mathbrown}{c}\color{mathbrown}{e}}\color{mathbrown}{\:}\mathrm{\color{mathbrown}{i}\color{mathbrown}{s}}\color{mathbrown}{\:} \\ $$ $$\color{mathbrown}{\mid}\mathrm{\color{mathbrown}{P}\color{mathbrown}{Q}}\color{mathbrown}{\mid}\color{mathbrown}{\:}\color{mathbrown}{=}\color{mathbrown}{\:}\sqrt{\color{mathbrown}{\left(}\mathrm{4}.\mathrm{0131}−\mathrm{1}.\mathrm{2067}\color{mathbrown}{\right)}^{\mathrm{\color{mathbrown}{2}}} \color{mathbrown}{+}\color{mathbrown}{\left(}\mathrm{0}.\mathrm{1145}−\mathrm{0}.\mathrm{7571}\color{mathbrown}{\right)}^{\mathrm{\color{mathbrown}{2}}} } \\ $$ $$\color{mathbrown}{\mid}\mathrm{\color{mathbrown}{P}\color{mathbrown}{Q}}\color{mathbrown}{\mid}\color{mathbrown}{\:}\color{mathbrown}{\approx}\color{mathbrown}{\:}\mathrm{2}.\mathrm{87903} \\ $$

Commented bybramlexs22 last updated on 09/Mar/21

y= x^3 −1 to x = y^2 + 4

$$\mathrm{y}=\:\mathrm{x}^{\mathrm{3}} −\mathrm{1}\:\mathrm{to}\:\mathrm{x}\:=\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{4} \\ $$

Commented bybramlexs22 last updated on 10/Mar/21

Commented bymr W last updated on 09/Mar/21

Edwin sir: please check, i think error is here: is same , we get { ((−(1/(3a^2 )) = −2b ; a^2 b = (1/6))),(((1/(3a))+a^3 −1 = 2b^3 +9b)) :}

$${Edwin}\:{sir}: \\ $$ $${please}\:{check},\:{i}\:{think}\:{error}\:{is}\:{here}: \\ $$ $$\mathrm{is}\:\mathrm{same}\:\color{mathred}{,}\color{mathred}{\:}\mathrm{we}\:\mathrm{get}\:\begin{cases}{−\frac{\mathrm{1}}{\mathrm{3a}^{\mathrm{\color{mathred}{2}}} }\:=\:−\mathrm{2b}\:;\:\mathrm{a}^{\mathrm{\color{mathred}{2}}} \mathrm{b}\:=\:\frac{\mathrm{1}}{\mathrm{6}}}\\{\frac{\mathrm{1}}{\mathrm{3a}}+\mathrm{a}^{\mathrm{3}} −\mathrm{1}\:=\:\mathrm{2b}^{\mathrm{3}} +\mathrm{9b}}\end{cases} \\ $$

Commented byEDWIN88 last updated on 09/Mar/21

yes sir. thanks you sir

$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by mr W last updated on 09/Mar/21

say point on curve y=x^3 −1 is P(p,p^3 −1) point on curve x=y^2 +4 is Q(q^2 +4,q) d=PQ D=d^2 =(q^2 +4−p)^2 +(q−p^3 +1)^2 (∂D/∂p)=−2(q^2 +4−p)−6p^2 (q−p^3 +1)=0 ⇒(q^2 +4−p)+3p^2 (q−p^3 +1)=0 ...(i) (∂D/∂q)=4q(q^2 +4−p)+2(q−p^3 +1)=0 ⇒2q(q^2 +4−p)+(q−p^3 +1)=0 ...(ii) ⇒2q=(1/(3p^2 )) ⇒q=(1/(6p^2 )) put this into (i): (1/(36p^4 ))+4−p+3p^2 ((1/(6p^2 ))−p^3 +1)=0 ⇒108p^9 −108p^6 +36p^5 −162p^4 −1=0 ⇒p≈1.2067 ⇒d_(min) ≈2.8791

$${say}\:{point}\:{on}\:{curve}\:{y}={x}^{\mathrm{3}} −\mathrm{1}\:{is} \\ $$ $${P}\left({p},{p}^{\mathrm{3}} −\mathrm{1}\right) \\ $$ $${point}\:{on}\:{curve}\:{x}={y}^{\mathrm{2}} +\mathrm{4}\:{is} \\ $$ $${Q}\left({q}^{\mathrm{2}} +\mathrm{4},{q}\right) \\ $$ $${d}={PQ} \\ $$ $${D}={d}^{\mathrm{2}} =\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)^{\mathrm{2}} +\left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$ $$\frac{\partial{D}}{\partial{p}}=−\mathrm{2}\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)−\mathrm{6}{p}^{\mathrm{2}} \left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$ $$\Rightarrow\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)+\mathrm{3}{p}^{\mathrm{2}} \left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0}\:\:\:...\left({i}\right) \\ $$ $$\frac{\partial{D}}{\partial{q}}=\mathrm{4}{q}\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)+\mathrm{2}\left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$ $$\Rightarrow\mathrm{2}{q}\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)+\left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0}\:\:\:...\left({ii}\right) \\ $$ $$\Rightarrow\mathrm{2}{q}=\frac{\mathrm{1}}{\mathrm{3}{p}^{\mathrm{2}} }\:\Rightarrow{\color{mathblue}{q}}\color{mathblue}{=}\frac{\mathrm{\color{mathblue}{1}}}{\mathrm{\color{mathblue}{6}}{\color{mathblue}{p}}^{\mathrm{\color{mathblue}{2}}} } \\ $$ $${put}\:{this}\:{into}\:\left({i}\right): \\ $$ $$\frac{\mathrm{1}}{\mathrm{36}{p}^{\mathrm{4}} }+\mathrm{4}−{p}+\mathrm{3}{p}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{6}{p}^{\mathrm{2}} }−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$ $$\Rightarrow\mathrm{108}{p}^{\mathrm{9}} −\mathrm{108}{p}^{\mathrm{6}} +\mathrm{36}{p}^{\mathrm{5}} −\mathrm{162}{p}^{\mathrm{4}} −\mathrm{1}=\mathrm{0} \\ $$ $$\Rightarrow{\color{mathblue}{p}}\color{mathblue}{\approx}\mathrm{\color{mathblue}{1}}\color{mathblue}{.}\mathrm{\color{mathblue}{2}\color{mathblue}{0}\color{mathblue}{6}\color{mathblue}{7}} \\ $$ $$\Rightarrow{\color{mathred}{d}}_{{\color{mathred}{m}\color{mathred}{i}\color{mathred}{n}}} \color{mathred}{\approx}\mathrm{\color{mathred}{2}}\color{mathred}{.}\mathrm{\color{mathred}{8}\color{mathred}{7}\color{mathred}{9}\color{mathred}{1}} \\ $$

Commented bymr W last updated on 09/Mar/21

Commented bybramlexs22 last updated on 09/Mar/21

thanks you sir

$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$

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