Question

Find the values of c such that the area of the region bounded by the parabolas y = 16x2 − c2 and y = c2 − 16x2 is 250/3. (enter your answers as a comma-separated list.)

Answer 1

First, note that these two parabolas are symmetric about the x-axis. The first one has its branches up in y-direction and the second one has its branches down in y-direction. This means that if you want to find area of the region bounded by the parabolas, you can find area of the region bounded by second parabola and x-axis and multiply it by 2.

1. Find x-intercepts of parabola $y=c^2-16x^2.$

When y=0, you have

$0=c^2-16x^2,\\ \\16x^2=c^2,\\ \\x_1=-\dfrac{c}{4},\ x_2=\dfrac{c}{4}.$

2. Find area:

$A_1= \int \limits_{-\frac{c}{4}}^{\frac{c}{4}}(c^2-16x^2)\, dx=\left(c^2x-16\dfrac{x^3}{3}\right)\big |^{\frac{c}{4}}_{-\frac{c}{4}}=\\ \\=c^2\left(\dfrac{c}{4}-\left(-\dfrac{c}{4}\right)\right)-\dfrac{16}{3}\left(\dfrac{c^3}{64}-\left(-\dfrac{c^3}{64}\right) \right)=\dfrac{c^3}{2}-\dfrac{c^3}{6}=\dfrac{c^3}{3}.$

Then the area of the region bounded by the parabolas is

$A=2A_1=2\cdot \dfrac{c^3}{3}.$

3. Since this area is $\dfrac{250}{3},$ you have

$2\cdot \dfrac{c^3}{3}=\dfrac{250}{3},\\ \\c^3=125,\\ \\c=5.$

Answer: c=5.