4 straight lines cross the origin in a coordinate plane, and all of them intersect a parabola whose equation is y = x^2 - 2 in 8 points.

Question

4 straight lines cross the origin in a coordinate plane, and all of them intersect a parabola whose equation is

y = x² - 2 in 8 points. What is the product of multiplying the x-coordinates of these 8 points?

a) only 16

b) only -16

c) only 8

d) only -8

e) there are multiple possible values

I know the answer is 16, and I can multiply possible values to get it, but why do we get 16 exactly? I suppose it has to do with the parabola properties, but I am not sure

Answer 1

You have to prove that the product is equal to 16.

y=x^2-2 is the parabola

y=mx is one of the lines because the line goes through the origin and thus b=0

set the right sides equal to each other

x^2-2=mx

x^2-mx-2=0

solve for x using the quadratic equation

[m±√(m^2+8)]/2

these are 2 of the 8 x-values

[m+√(m^2+8)]/2 and [m-√(m^2+8)]/2

now multiply these two values together and you get...

[m^2-(m^2+8)]/4

(m^2-m^2-8)/4

-8/4=-2

no matter what the slope "m" is, "m^2" gets cancelled out and you are always left with -2

the second line, no matter what it is, no matter what the slope is, will yield another -2 (don't forget that the lines always go through the origin)

the third line gives another -2

the fourth line gives another -2

you get -2 four times for the four x-values

(-2)(-2)(-2)(-2)=16

Answer 2

The straight lines must have equations in the form y = mx.

1. )

Assume m = 1

Line and parabola intersect at x = x² - 2

0 = x² - x - 2

0 = (x - 2)(x + 1)

The two x values are 2 and -1

Assume m = -1

-x = x² - 2

0 = x² + x - 2

0 = (x + 2)(x -1)

The two x values are -2 and 1.

Assume m = 2

2x = x² - 2

0 = x² - 2x - 2

The two x values are 1 ± √3

Assume m = -2

-2x = x² - 2

0 = x² + 2x - 2

The two x values are -1 ± √3

The product of the four x values is 16.

This is not a general proof. It is only a demonstration.