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# Slope of line through origin that divides region bounded by y = 8x-2x^2 & the x-axis into 2 equal area regions?

What is the slope of the line that passes through the origin and divides the region bounded by the parabola y = 8x - 2x^2 and the x-axis into two regions with equal area.

I tried those computations and got roughly m = 2.7284731, however this is incorrect.

### 1 Answer

- Anonymous1 decade agoFavorite Answer
It will help if you draw a rough graph.

The parabola is y = -2x(x-4), so it crosses the x-axis at (0,0) and again at (4,0). So to first find the total area under the parabola, integrate ∫ 8x - 2x^2 dx from x=0 to 4.

A line that passes through the origin has an equation of y=mx. We just need to find the value of m such that the line divides the original area into two pieces of equal area.

The line intersects the parabola when

mx = 8x - 2x^2

2x^2 + mx - 8x = 0

2x^2 + (m-8)x = 0

x (2x + (m-8)) = 0

So this happens when x=0 (which we already know; both the parabola and the line pass through the origin), and when 2x+m-8=0, which is when x=(8-m)/2.

Consider the area bounded on the top left by the line y=mx, on the top right by the parabola from x=(8-m)/2 to x=4, and on the bottom by the x-axis. If you draw a vertical line through x=(8-m)/2, this splits the area up into a triangular piece on the left and a curved piece on the right. The area of this is

∫ mx dx + ∫ 8x-2x^2 dx

where the first integral is taken from x=0 to x=(8-m)/2, an the second integral is taken from x=(8-m)/2 to x=4.

Calculate this. Now set it equal to HALF of the total area you found earlier. Solve for m.