Area bounded by a parabola and its chord is equal to two-thirds of area of bounding paralellogram

Question

Here AB|| CD and ABCD is a parallelogram . We need to show that the area covered by the both Parallelogram and the Parabola CDX is two thirds of that of the parallelogram ABCD.

Can anyone please help me to prove this?

Answer 1

Let $y=ax^2$ be the parabola, with $(x_1, ax_1^2)$ and $(x_2, ax_2^2)$ the coordinates of the chord. The equation of the chord is then

$$y=kx+m_1=a(x_1+x_2)x - ax_1x_2$$

Let the side of the parallelogram parallel to the chord $y=kx +m_2$. Since it is tangential to the parabola, the discriminant of $a(x_1+x_2)x +m_2=ax^2$ is zero, which yield $m_2=-\frac{k^2}{4a}$.

The area of the enclosing parallelogram is then,

$$A_1 = (m_1-m_2)(x_2-x_1) = \left(- ax_1x_2+\frac{a^2(x_1+x_2)^2}{4a}\right)(x_2-x_1) =\frac a4(x_2-x_1)^3\tag{1}$$

The area of the parabolic segment can be obtained with,

$$A_2 = \frac a2(x_1^2+x_2^2)(x_2-x_1)-\int_{x_1}^{x_2} ax^2 dx=\frac a6(x_2-x_1)^3\tag{2}$$

where the first term is the area of the trapezoid under the chord and the integral accounts for the area under the parabola.

From (1) and (2), $A_2 = \frac23 A_1$, that is, the area bounded by the parabola and its chord is equal to two-thirds of the area of the bounding parallelogram.