Proving that the Area of a Parabolic Segment is Two-Thirds of the Circumscribed Rectangle

In this article, we aim to prove that the area of the segment of the parabola defined by the equation (y^2 frac{c^2}{d} x) cut off by the vertical line (x d) is two-thirds of the area of the rectangle circumscribed about its segment. This proof involves a series of steps including finding the points of intersection, determining the area of the parabolic segment, and the area of the circumscribed rectangle.

Step 1: Finding the Points of Intersection

The parabola (y^2 frac{c^2}{d} x) intersects the vertical line (x d) at:

(y^2 frac{c^2}{d} d c^2 implies y pm c)

Therefore, the points of intersection are (d c) and (d -c).

Step 2: Determining the Area of the Segment

The area (A) of the segment of the parabola cut off by (x d) can be calculated using integration. The area under the parabola from (x 0) to (x d) is:

(A int_{0}^{d} y , dx)

Since (y sqrt{frac{c^2}{d} x}), we have:

(A int_{0}^{d} sqrt{frac{c^2}{d} x} , dx sqrt{frac{c^2}{d}} int_{0}^{d} sqrt{x} , dx)

Calculating the integral:

(int_{0}^{d} sqrt{x} , dx left[frac{2}{3} x^{frac{3}{2}}right]_{0}^{d} frac{2}{3} d^{frac{3}{2}})

Thus the area becomes:

(A sqrt{frac{c^2}{d}} cdot frac{2}{3} d^{frac{3}{2}} frac{2c}{3} d)

Step 3: Finding the Area of the Circumscribed Rectangle

The circumscribed rectangle has a width of (d) and a height of (2c) since it stretches from (-c) to (c). Therefore, the area (A_R) of the rectangle is:

(A_R text{width} times text{height} d times 2c 2cd)

Step 4: Relating the Areas

We have found:

The area of the segment (A frac{2c}{3} d) The area of the rectangle (A_R 2cd)

To show that the area of the segment is two-thirds of the area of the rectangle, we compute:

(frac{A}{A_R} frac{frac{2c}{3} d}{2cd} frac{2c}{3} cdot frac{1}{2c} frac{1}{3})

Thus we find:

(A frac{2}{3} A_R)

Conclusion

The area of the segment of the parabola (y^2 frac{c^2}{d} x) cut off by (x d) is indeed two-thirds of the area of the rectangle circumscribed about its segment. This completes the proof.

For [y^2 leftfrac{c^2}{d}right x] when (x d), then [y^2 leftfrac{c^2}{d}right d c^2] [y pm c]

So the vertical line (x d) intersects the parabola at points [d c text{ and } d -c]

Now we'll draw a graph showing the segment of the parabola cut off by the vertical line as well as the circumscribed rectangle.

The circumscribed rectangle has width ( d) and height ( 2c)

[textbf{Area_{rectangle}} d cdot 2c 2cd]

To find the area bounded by the parabola and the vertical line, we integrate with respect to (y) on the interval (-c le y le c). So we rewrite the equation of the parabola as:

[x frac{d}{c^2} y^2]

The region is bounded to the left by the parabola and to the right by the vertical line (x d)

[textbf{Area_{parabola}} int_{-c}^{c} left(d - frac{d}{c^2}y^2right)dy]

[ left[dy - frac{d}{3c^2}y^3right]_{-c}^{c}]

[ dleft(c - (-c)right) - frac{d}{3c^2}left(c^3 - (-c^3)right)]

[ dleft(2cright) - frac{d}{3c^2}left(2c^3right)]

[ 2cd - frac{2}{3}cd]

[ frac{4}{3}cd frac{2}{3}left(2cdright)]

[ textbf{frac{2}{3} Area_{rectangle}}]

This confirms our initial proof that the area of the segment is two-thirds of the area of the circumscribed rectangle.