Parabola and point

Calculus Level 4

Let the shortest distance between the point ( 2 , 5 ) (2,5) and the parabola y = x 2 y=x^2 be

1 2 ⋅ a − 4 b b , \frac 12 \cdot \sqrt{a-4b\sqrt{b}},

where a a and b b are coprime. Find a + b . a+b.

Bonus: Generalize this for the point ( h , h 2 + 1 ) . \big(h,h^2+1\big).


The answer is 65.

Mrigank Shekhar Pathak by Mrigank Shekhar Pathak , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

2 solutions

For the generalization the sum is also 65, as shown in image of my worked-out solution.

2 Helpful 0 Interesting 0 Brilliant 2 Confused
Hitesh Yadav
Jan 11, 2021

We can use the property that shortest distance from a point to a line is perpendicular . Now a parabolic curve can be understood as infinitely short and infinite number of lines . So the shortest distance would have to be a normal to point on parabola passing through (2,5) By solving the cubic equation t 3 − 18 t + 8 = 0 t^{3}-18t+8=0 We find that t can have three possible values . One of them using factor theorem is 4 (thank god!). Other two by solving the quadratic. Now draw a rough sketch of quadratic function and you can possibly deduce that x has to be equal to ( 2 + 6 ) / 2 (2+\sqrt{6})/2 and then rest can be done quite easily.

0 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...