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Question Number 123243 by benjo_mathlover last updated on 24/Nov/20
 What is the shortest distance  between two parabolas   y^2  = x−2 and x^2  = y−2 .
Whatistheshortestdistancebetweentwoparabolasy2=x2andx2=y2.
Commented by liberty last updated on 24/Nov/20
In other way by normal line of two  parabolas :y=x^2 +2 { ((grad normal = −(1/(2x)))),((point P(a,a^2 +2))) :}  :y^2  =x−2  { ((grad normal =−2y)),((point Q(b^2 +2,b) )) :}   { ((equation of normal ⇒y_1 =−(1/(2a))x+a^2 +2)),((equation of normal ⇒y_2 =−2bx+2b^3 +5b)) :}  these are the same line .   { ((−(1/(2a)) = −2b ; a=(1/(4b)))),((a^2 +2 = 2b^3 +5b ; 64b^5 +160b^3 −80b^2 −2=0)) :}  The only real number answer is   b=(1/2) and a=(1/2) so the two points  on the parabolas are  { ((P((1/2),(9/4)))),((Q((9/4),(1/2)))) :}  The distance = (√(2×((9/4)−(2/4))^2 ))                                = (7/4)(√2)
Inotherwaybynormallineoftwoparabolas:y=x2+2{gradnormal=12xpointP(a,a2+2):y2=x2{gradnormal=2ypointQ(b2+2,b){equationofnormaly1=12ax+a2+2equationofnormaly2=2bx+2b3+5bthesearethesameline.{12a=2b;a=14ba2+2=2b3+5b;64b5+160b380b22=0Theonlyrealnumberanswerisb=12anda=12sothetwopointsontheparabolasare{P(12,94)Q(94,12)Thedistance=2×(9424)2=742
Commented by benjo_mathlover last updated on 24/Nov/20
Commented by benjo_mathlover last updated on 24/Nov/20
correct. thank you
correct.thankyou
Answered by bobhans last updated on 24/Nov/20
  (•) y^2 =x−2              (•) x^2 =y−2    2y.y′ = 1                           2x = y′   grad y′=(1/(2y))=(1/(2a))                                 grad y′=2x=2b    Point A(a^2 +2,a)                   Point B(b,b^2 +2)  tangent line                         tangent line  y=(1/(2a))(x−(a^2 +2))+a      y=2b(x−b)+b^2 +2  two tangent line are parallel  ⇒ (1/(2a)) = 2b ⇔ b = (1/(4a))  so the shortest distance is  ∣AB∣ =(√((a^2 +2−b)^2 +(a−(b^2 +2))^2 ))  =(√((a^2 +2−(1/(4a)))^2 +(a−(1/(16a^2 ))−2)^2 ))  =(√((((4a^3 +8a−1)/(4a)))^2 +(((16a^3 −32a^2 −1)/(16a^2 )))^2 ))  we compute ((d∣AB∣)/da) = 0  we get a=(1/2) ∧ b=(1/2)  ∣AB∣_(min)  = (√(((1/4)+2−(1/2))^2 +((1/2)−(1/4)−2)^2 ))   = (√(2((9/4)−(2/4))^2 )) = ((7(√2))/4) .
()y2=x2()x2=y22y.y=12x=ygrady=12y=12agrady=2x=2bPointA(a2+2,a)PointB(b,b2+2)tangentlinetangentliney=12a(x(a2+2))+ay=2b(xb)+b2+2twotangentlineareparallel12a=2bb=14asotheshortestdistanceisAB=(a2+2b)2+(a(b2+2))2=(a2+214a)2+(a116a22)2=(4a3+8a14a)2+(16a332a2116a2)2wecomputedABda=0wegeta=12b=12ABmin=(14+212)2+(12142)2=2(9424)2=724.

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