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 .](https://www.tinkutara.com/question/Q123243.png)
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)](https://www.tinkutara.com/question/Q123245.png)
Commented by benjo_mathlover last updated on 24/Nov/20
![](https://www.tinkutara.com/question/17154.png)
Commented by benjo_mathlover last updated on 24/Nov/20
![correct. thank you](https://www.tinkutara.com/question/Q123247.png)
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) .](https://www.tinkutara.com/question/Q123244.png)