1421. Distance between two parabolae

    Pepe Chapuzas had drawn two parabolae that didn't cut each other on the blackboard. One was convex and the other concave... It was a challenge...

    Dear Teacher:
    What's the distance between the parabolae if their equations are

f(x)  =  x²/3 + 47/3

g(x)  =  − x²/6 + 10x/3

    Susi Fogata did the calculations...

    Dear Teacher:
    These parabolae don't cut each other because the equation

 x²/3 + 47/3  =  − x²/6 + 10x/3

or better

3x² − 20x + 94  =  0

hasn't any solutions because its discriminant

20² − 4·3·94  =  − 728  <  0

    If  S (s, s²/3 + 47/3)  is a point of the parabola  f  and  T (t, − t²/6 + 10t/3)  is a point of the parabola  g , and are the nearest ones, then the slopes of  f  and  g  at  S  and  T  are equal...  

 f ' (s)  =  g ' (t)
2s/3  =  − t/3 + 10/3
t  =  10 − 2s

    On the other hand, the vector  TS  is perpendicular to  f  at  S ...

(s − t, f(s) − g(t)) · (1, f ' (s))  =  0
(s − t, s²/3 + 47/3 + t²/6 − 10t/3) · (1, 2s/3)  =  0
(s − (10−2s), s²/3 + 47/3 + (10−2s)²/6 − 10(10−2s)/3) · (1, 2s/3)  =  0
(3s − 10, s²/3 + 47/3 + 50/3 − 20s/3 + 2s²/3 − 100/3 + 20s/3) · (1, 2s/3)  =  0
(3s − 10, s² − 1) ·  (1, 2s/3)  =  0
3s − 10 + 2s³/3 − 2s/3  =  0
 2s³ + 7s − 30  =  0

    s = 2  is the only solution because  4² − 4·2·15  =  − 104  <  0 .

    So, we have the points  S (2, 17)  and  T (6, 14)  and the vector  TS = (−4, 3) , hence the distance is  |TS| = √(16+9) = √25 = 5 .