Archimedes' rule that area of a parabola is 4/3 times an inscribed triangle

[runner07]

Homework Statement

Let ABC be a piece of a parabola. The point B is chosen such that the tangent to the parabola at B is parallel to the line AC. Archimedes proved the Area of parabola inside ABC is 4/3 times the triangle ABC. Prove this using calculus

Homework Equations

The integral from a to c (Where pt A=(a,a^2), pt B=(b,b^2), and pt C=(c,c^2)) of (x^2)-(x+c)

The Attempt at a Solution

I have been trying to figure this out for hours. I know you can assume the parabola to be in the standard form y=x^2 without loss of generality because the problem relies on ratios of areas. And from this, I thought you could assume the segment AC has standard form y=x+c. Then, since we are looking for area, we could use the method of area between two curves, which is why I feel like the integral from a to c of (x+c)-(x^2) would be right. But when I calculate this, it should end up looking something like 4/3(base*height*1/2) and it definitely doesn't look that way. Any help would be greatly appreciated!

 

[mighty2000]

Thank you for your question. I will try my best to guide you through the solution using calculus.

First, let's define the parabola as y = x^2, where the focus is at the origin (0,0) and the directrix is the line y = -1. Let's also assume that the points A, B, and C have coordinates (a,a^2), (b,b^2), and (c,c^2) respectively.

Since the tangent to the parabola at B is parallel to the line AC, we can say that the slope of the tangent at B is equal to the slope of the line AC. Using the slope formula, we have:

m = (b^2 - a^2)/(b - a) = (c^2 - a^2)/(c - a)

We can simplify this to get:

b + a = c + a

Therefore, b = c.

Now, let's find the equation of the tangent line at B. We know that the slope of the tangent at any point on the parabola is given by the derivative of the parabola at that point. So, the slope of the tangent at B is given by:

m = dy/dx = 2b

Since we know that b = c, we can rewrite this as:

m = 2c

Using the point-slope form of a line, we can write the equation of the tangent line at B as:

y - b^2 = 2c(x - b)

Now, let's find the intersection points of this tangent line and the parabola. We can substitute the equation of the tangent line into the equation of the parabola to get:

x^2 = 2cx - c^2 + b^2

Since b = c, we can simplify this to:

x^2 = 2cx - c^2 + c^2

x^2 = 2cx

This is a quadratic equation with two solutions: x = 0 and x = 2c. Since we are only interested in the portion of the parabola between points A and C, we can ignore the solution x = 0. Therefore, the x-coordinate of point C is 2c.

Now, we can use the method of area between two curves to find the area of the parabola inside the triangle ABC. The area is