   Chapter 4.P, Problem 14P

Chapter
Section
Textbook Problem

# The figure shows a parabolic segment, that is, a portion of a parabola cut off by a chord AB. It also shows a point C on the parabola with the property that the tangent line at C is parallel to the chord AB. Archimedes proved that the area of the parabolic segment is 4 3 times the area of the inscribed triangle ABC. Verify Archimedes’ result for the parabola y = 4 − x 2 and the line y = x + 2 . To determine

To prove:

The area of parabola segment is 43 times the area of the inscribed triangle ABC.

Explanation

1) Concept:

The coordinates of point C: Place the equation of line in the parabola, and the intersecting points will give us C coordinates.

Then find the area of parabola segment =Area under the parabola Area of triangle ABD(area under the line)

Then calculate the ratio of area of parabola segment ACB to the area of the triangle.

2) Calculation:

Let the line parallel to y=x+2 is y=x+k

Since it touches the parabola, it will satisfy the equation of the parabola at a point c.

x+k=4-x2

Add x2-4 on both sides, and simplify.

x2+x+k-4=0

Discriminant D for the above equation should be zero because the line is tangent to the parabola.

D=b2-4ac=0

Here, a=1, b=1, c=k-4

Substitute above values in D.

D=12-41k-4=0

Simplify.

k=174

So y=x+k=x+174

To get the coordinates of point C, substitute the equation of line in the parabola, and the intersecting points will give us the C coordinates.

x+174=4-x2

Add x2-4 to both the sides, and simplify.

x2+x+14=0

Solve the above equation.

x=-12

Substitute the above value in y=x+174.y=-12+174

Simplify.

y=154

Therefore, the coordinates of C are -12, 154. To calculate points of intersection of the given line and parabola,

x+2=4-x2

Add x2-4 on both sides, and simplify.

x2+x-2=0

Simplify.

x=-2, 1

Put x=-2, 1 in y=x+2

y=-2+2=0 And

y=1+2=3

Now the area of the parabola segment = Area under parabola Area of triangle ABD (area under line). Thus

A=-21(4-x2)dx--21(x+2)dx

Integrating

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