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Question Number 134040 by benjo_mathlover last updated on 27/Feb/21
Find the shortest distance between the curve y^2 = x−1and x^2 = y−1
$$\mathrm{Find}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance}\: \\ $$$$\mathrm{between}\:\mathrm{the}\:\mathrm{curve}\: \\ $$$$\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{x}−\mathrm{1and}\:\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}−\mathrm{1} \\ $$
Commented by benjo_mathlover last updated on 27/Feb/21
Answered by EDWIN88 last updated on 27/Feb/21
(i) gradient of tangent line curve y^2 =x−1 at point A(a^2 +1, a) is m_1 = (1/(2(√(a^2 +1−1))))=(1/(2a)) then gradient of normal line is −2a⇒equation of normal line :y=−2a(x−a^2 −1)+a y=−2ax+2a^3 +3a (2)gradient of tangent line curve y=x^2 +1 at point B(b,b^2 +1) is m_2 = 2b then gradient of normal line is −(1/(2b)) ⇒equation of normal line : y=−(1/(2b))(x−b)+b^2 +1 y = −(1/(2b))x+ b^2 +(3/2) Thus the equation of normal line is same we get { ((2a=(1/(2b)) ; a=(1/(4b)))),((2a^3 +3a = b^2 +(3/2))) :} ⇔ (2/(64b^3 ))+(3/(4b)) = ((2b^2 +3)/2) ⇒((2+48b^2 )/(64b^3 )) = ((2b^2 +3)/2) ⇒ 2+48b^2 = 32b^3 (2b^2 +3) ⇒64b^5 +96b^3 −48b^2 −2=0 ⇒32b^5 +48b^3 −24b^2 −1=0 ⇒(2b−1)(16b^4 +8b^3 +28b^2 +2b+1)=0 { ((b=(1/2) ⇒B((1/2), (5/4)))),((a=(1/2)⇒A((5/4),(1/2)))) :} ⇒AB_(min) = (√(2((5/4)−(1/2))^2 ))=(3/4)(√2)
$$\left(\mathrm{i}\right)\:\mathrm{gradient}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\:\:\mathrm{curve}\:\mathrm{y}^{\mathrm{2}} =\mathrm{x}−\mathrm{1} \\ $$$$\mathrm{at}\:\mathrm{point}\:\mathrm{A}\left(\mathrm{a}^{\mathrm{2}} +\mathrm{1},\:\mathrm{a}\right)\:\mathrm{is}\:\mathrm{m}_{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}}=\frac{\mathrm{1}}{\mathrm{2a}}\:\mathrm{then}\: \\ $$$$\mathrm{gradient}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\:\mathrm{is}\:−\mathrm{2a}\Rightarrow\mathrm{equation} \\ $$$$\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\::\mathrm{y}=−\mathrm{2a}\left(\mathrm{x}−\mathrm{a}^{\mathrm{2}} −\mathrm{1}\right)+\mathrm{a} \\ $$$$\:\mathrm{y}=−\mathrm{2ax}+\mathrm{2a}^{\mathrm{3}} +\mathrm{3a} \\ $$$$\left(\mathrm{2}\right)\mathrm{gradient}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{2}} +\mathrm{1} \\ $$$$\mathrm{at}\:\mathrm{point}\:\mathrm{B}\left(\mathrm{b},\mathrm{b}^{\mathrm{2}} +\mathrm{1}\right)\:\mathrm{is}\:\mathrm{m}_{\mathrm{2}} =\:\mathrm{2b}\:\mathrm{then}\: \\ $$$$\mathrm{gradient}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\:\mathrm{is}\:−\frac{\mathrm{1}}{\mathrm{2b}}\:\Rightarrow\mathrm{equation} \\ $$$$\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\::\:\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{2b}}\left(\mathrm{x}−\mathrm{b}\right)+\mathrm{b}^{\mathrm{2}} +\mathrm{1} \\ $$$$\:\mathrm{y}\:=\:−\frac{\mathrm{1}}{\mathrm{2b}}\mathrm{x}+\:\mathrm{b}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathrm{Thus}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\:\mathrm{is}\:\mathrm{same} \\ $$$$\mathrm{we}\:\mathrm{get}\:\begin{cases}{\mathrm{2a}=\frac{\mathrm{1}}{\mathrm{2b}}\:;\:\mathrm{a}=\frac{\mathrm{1}}{\mathrm{4b}}}\\{\mathrm{2a}^{\mathrm{3}} +\mathrm{3a}\:=\:\mathrm{b}^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{2}}}\end{cases} \\ $$$$\Leftrightarrow\:\frac{\mathrm{2}}{\mathrm{64b}^{\mathrm{3}} }+\frac{\mathrm{3}}{\mathrm{4b}}\:=\:\frac{\mathrm{2b}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}}\:\Rightarrow\frac{\mathrm{2}+\mathrm{48b}^{\mathrm{2}} }{\mathrm{64b}^{\mathrm{3}} }\:=\:\frac{\mathrm{2b}^{\mathrm{2}} +\mathrm{3}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{2}+\mathrm{48b}^{\mathrm{2}} \:=\:\mathrm{32b}^{\mathrm{3}} \left(\mathrm{2b}^{\mathrm{2}} +\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{64b}^{\mathrm{5}} +\mathrm{96b}^{\mathrm{3}} −\mathrm{48b}^{\mathrm{2}} −\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{32b}^{\mathrm{5}} +\mathrm{48b}^{\mathrm{3}} −\mathrm{24b}^{\mathrm{2}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{2b}−\mathrm{1}\right)\left(\mathrm{16b}^{\mathrm{4}} +\mathrm{8b}^{\mathrm{3}} +\mathrm{28b}^{\mathrm{2}} +\mathrm{2b}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\begin{cases}{\mathrm{b}=\frac{\mathrm{1}}{\mathrm{2}}\:\Rightarrow\mathrm{B}\left(\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{5}}{\mathrm{4}}\right)}\\{\mathrm{a}=\frac{\mathrm{1}}{\mathrm{2}}\Rightarrow\mathrm{A}\left(\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right)}\end{cases}\:\Rightarrow\mathrm{AB}_{\mathrm{min}} \:=\:\sqrt{\mathrm{2}\left(\frac{\mathrm{5}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }=\frac{\mathrm{3}}{\mathrm{4}}\sqrt{\mathrm{2}} \\ $$
Answered by MJS_new last updated on 27/Feb/21
symmetry about y=x it′s very easy to solve...
$$\mathrm{symmetry}\:\mathrm{about}\:{y}={x} \\ $$$$\mathrm{it}'\mathrm{s}\:\mathrm{very}\:\mathrm{easy}\:\mathrm{to}\:\mathrm{solve}… \\ $$
Answered by benjo_mathlover last updated on 27/Feb/21
since both curve symmetry to line y = x , gradient of normal line curve y^2 =x−1⇒ −2(√(x−1)) =−1 ⇒(√(x−1)) =(1/2) ; x=(5/4) ∧ y = (1/2) we get point A((5/4),(1/2)) gradient of normal line curve y = x^2 +1⇒−(1/(2x)) = −1 ; x=(1/2)∧ y=(5/4) we get point B((1/2), (5/4)). So the shortest distance between the given two curve is AB = (√(((5/4)−(1/2))^2 +((1/2)−(5/4))^2 )) = (√(2×((3/4))^2 )) = ((3(√2))/4)
$$\mathrm{since}\:\mathrm{both}\:\mathrm{curve}\:\mathrm{symmetry}\:\mathrm{to}\:\mathrm{line}\: \\ $$$$\mathrm{y}\:=\:\mathrm{x}\:,\:\mathrm{gradient}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line} \\ $$$$\mathrm{curve}\:\mathrm{y}^{\mathrm{2}} =\mathrm{x}−\mathrm{1}\Rightarrow\:−\mathrm{2}\sqrt{\mathrm{x}−\mathrm{1}}\:=−\mathrm{1} \\ $$$$\Rightarrow\sqrt{\mathrm{x}−\mathrm{1}}\:=\frac{\mathrm{1}}{\mathrm{2}}\:;\:\mathrm{x}=\frac{\mathrm{5}}{\mathrm{4}}\:\wedge\:\mathrm{y}\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{point}\:\mathrm{A}\left(\frac{\mathrm{5}}{\mathrm{4}},\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\mathrm{gradient}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\:\mathrm{curve} \\ $$$$\mathrm{y}\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{1}\Rightarrow−\frac{\mathrm{1}}{\mathrm{2x}}\:=\:−\mathrm{1}\:;\:\mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}}\wedge\:\mathrm{y}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{point}\:\mathrm{B}\left(\frac{\mathrm{1}}{\mathrm{2}},\:\frac{\mathrm{5}}{\mathrm{4}}\right). \\ $$$$\mathrm{So}\:\mathrm{the}\:\mathrm{shortest}\:\mathrm{distance}\:\mathrm{between} \\ $$$$\mathrm{the}\:\mathrm{given}\:\mathrm{two}\:\mathrm{curve}\:\mathrm{is} \\ $$$$\mathrm{AB}\:=\:\sqrt{\left(\frac{\mathrm{5}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{4}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:=\:\sqrt{\mathrm{2}×\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{2}} }\:=\:\frac{\mathrm{3}\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$ \\ $$
Commented by benjo_mathlover last updated on 27/Feb/21

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