Parabolas y=−2x2 and y=2x2+k intersect at points A and B that are in the third and the fourth quadrants respectively. Find k if length of the segment AB is 5.

Question

Parabolas y=−2x² and y=2x²+k intersect at points A and B that are in the third and the fourth quadrants respectively. Find k if length of the segment AB is 5. 

Answer 1

Hi Jack K.,

Parabola (1) y = -2x² is a parabola that opens downward with the vertex at (0, 0).

Parabola (2) y = 2x² + k is a parabola that opens up with the vertex at (0, k).

We will need to find k, so that the two intersections (A and B) of these two parabolas is a distance of 5 units apart. With the intersection in the III and IV quadrant, parabola (2) will need the vertex to shift down on the y -axis.

The distance of the intersection will be a horizontal line from A to B. So the distance is a change in x-value with no change in y-value (|x₂ - x₁| = |B_x - A_x| = distance).

Since both parabolas have their vertex on the y-axis, half the distance will be in quadrant III, and the other half of the distance in quadrant IV. So the y-axis splits the distance, and since the distance is only values of x, half the distance is 5/2 = 2.5. We now know that the intersection will be at (-2.5, y) and

(2.5, y).

Also the y-values at the intersection will have to be equal. We can set the two equations equal to each other (y = y), and with x = to 2.5 or -2.5, we can solve for k.

-2x² = 2x² + k

-25 = k, (I'll let you do the math)

Therefore y = 2x² - 25, is a shift in the parabolas vertex down to -25 on the y-axis with the intersection of the two parabolas at (-2.5, -12.5) and (2.5, -12.5).

I hope this helps, Joe.

Answer 2

Let’s say

f(x)= -2x² and g(x)= 2x²+k

Let A and B be their intersection points such that;

A= (x₁,y₁) and B= (x₂,y₂)

Both functions are parabolas.

According to transformations, f(x) opens down and has a vertex at 0.

According to transformations, g(x) is very similar to f(x) except it opens up and is shifted vertically (because of k).

There is no horizontal shift in either function so at their intersections, A and B, the y-values are the same (since they are parabolas).

So it follows that;

A= (x₁,y) and B= (x₂,y)

Because there is no horizontal shift, both functions are symmetric about the y-axis.

This means that if two x’s have the same y-value, they are the positive and negative versions of each other.

So it follows that;

i) x₁ = -x₂

Since the distance between A and B is 5 and the line connecting them is perfectly horizontal (since the y values are the same), this means that the distance between x₁ and x₂ is 5.

Thus,

ii) x₂ - x₁ = 5

Using equations (i) and (ii),

x₁ = -x₂ x₂ - x₁ = 5

solve for one of the variables

x₁ = -x₂ x₁ = x₂ - 5

set each equation equal to each other and solve

-x₂ = x₂ - 5

x₂ = 2.5

since x₂ and x₁ are opposite signed,

x₁ = -2.5

Now that we have the x-values, let’s find k.

Since an intersection is where f(x) = g(x) we can plug in either of our x-values into this equation.

f(x) = g(x)

-2x² = 2x² + k

Plug in x₁ or x₂

-2(2.5)² = 2(2.5)² + k

-12.5 = 12.5 + k

k = -25