1958 AMC 12 Problem 22

Question

A particle is placed on the parabola

y=x^2-x-6

at a point

P

whose ordinate is

6.

It is allowed to roll along the parabola until it reaches the nearest point

Q

whose ordinate is

-6.

The horizontal distance traveled by the particle (the numerical value of the difference in the abscissas of

P

and

Q

) is:

5

4

3

2

1

Find the two

x

-coordinates where the parabola has ordinate

6

and the two where it has ordinate

-6

Pair each upper point with the nearer lower point on the same branch

For

y=6,

x^2-x-12=0,

so

x=4

or

-3.

For

y=-6,

x^2-x=0,

so

x=0

or

1.

The nearest lower point to

x=4

is

x=1,

and the nearest to

x=-3

is

x=0.

Either horizontal distance is

3.

Therefore, the correct answer is C.

Problem 22 in other years