There is a line through the origin that divides the region bounded by the parabola y = 6x-5x^2 and the x-axis into two regions with equal area. What is the slope of that line?

Answer 1

Hi! Good question.

So first make a graph of the parabola just to get a picture of what is going on.

Now we are told that the line divides the area in half so lets try to convert that to math shorthand.

Let's first find the total area between the parabola and the x axis.

Find where the parabola intersects the x axis. (6x-5x²=0)

This occurs when x=0 and x=1.2

Now, find the area:

∫₀^1.26x-5x² dx = total area

This comes out to be 1.44

Now we know the line (of form y=mx+b) must go through the point (0,0), this lets us know that b=0.

Now, we have y=mx

We know that the line y=mx intersects the parabola at some point.

This occurs when mx=6x-5x²

Solve for the x value.

This comes out to be when x=(6-m)/5

Now we know that the area between our line and the x axis from 0 to (6-m)/5 plus the area between the parabola and the x axis from (6-m)/5 to 1.2 must equal 1.44/2=0.72

(draw a picture of this situation if you are confused!)

So, lets write out those integrals!

∫₀^(6-m)/5 mx dx + ∫^1.2_(6-m)/5 6x-5x² dx = 0.72

just solve all this out for 'm' and you have your answer!

(should be around 1.2378 ish)

Answer 2

Let equation of this line y = mx. Area between parabola y = 6x - 5x² and x-axis: x-intercepts, y = 0,

6x - 5x² = 0, x = 0 and x = 6/5 = 1.2;

A = ∫₀^1.2(6x - 5x²)dx = (3x² - 5x³/3)₀^1.2 = 3·1.2² - 5·1.2³/3 = 1.44.

Now x-coordinate of point of intersection y = mx and parabola y = 6x - 5x²; mx = 6x - 5x²; x = 0 and

x = (6 - m)/5. Now aea betwween parabola and line A/2 = ∫₀^(6-m)/5(6x - 5x² - mx)dx = (3x² - 5x³/3 - mx²/2)₀^{(6 - m)/5} = ((6 - m)/5)²(3 - m/2 - 5/3(6 - m)/5) = (6 - m)²/25(1 - m/6) = (6 - m)³/150 = 0.72;

(6 - m)³ = 108; 6 - m = (108)^1/3, m = 6 - (108)^1/3 = 6 - 3(4)^1/3.

Slope m = 6 - 3(4)^1/3

Answer 3

Hi Shreeya!

You are looking to set up and evaluate two integrals, which in turn must be equal to each other. Let's think about how to do that.

First, draw a picture of the parabola. You can write the formula for it as y = 6x - 5x^2 or x*(6 - 5x). y will be zero at x=0 and at x = 6/5. One of your integrals will go between these values of x, and the other will only go between x = 0 and the value of x at which the parabola meets the line y = ax, with a the unknown slope.

Now, think of each integral as a sum of toothpicks that go from the low y to the high y across each space. One integral will be between the parabola and the line, and the other will be between the unknown line and the x axis up to the point where the parabola intersects the unknown line, and between the parabola and the x axis for values of x between the intersection x and x = 6/5. You can use what you know of integration to solve for each integral in terms of a. Setting the two integrals equal to each other will help you solve for a and get an answer to your problem.

Cheers,

Joe