Question

Find the values of c such that the area of the region bounded by the parabolas

y = 16x^2 − c^2 and y = c^2 − 16x^2
is 144

Answer 1

For there to be a region bounded by the two parabolas, you first need to find some conditions on
`c`. The two parabolas must intersect each other twice, so you need two solutions to


`16x^2-c^2=c^2-16x^2`

You have


`32x^2=2c^2\implies 16x^2=c^2\implies x=\pm\frac4`

which means you only need to require that
`c\\eq0`. With that, the area of any such bounded region would be given by the integral


`\displaystyle\int_(-4|c|)^(4|c|)\bigg((c^2-16x^2)-(16x^2-c^2)\bigg)\,\mathrm dx`

since
`c^2-16x^2>16x^2-c^2` for all
`c\\eq0`. Now,


`\displaystyle\int_(-|c|/4)^(|c|/4)\bigg((c^2-16x^2)-(16x^2-c^2)\bigg)\,\mathrm dx=2\int_0^(|c|/4)(2c^2-32x^2)\,\mathrm dx`

by symmetry across the y-axis. Integrating yields


`\displaystyle2\int_0^(|c|/4)(2c^2-32x^2)\,\mathrm dx=4\int_0^(|c|/4)(c^2-16x^2)\,\mathrm dx`

`=4\left[c^2x-\frac{16}3x^3\right]_(x=0)^(x=|c|/4)`

`=c^2|c|-\fracc3`

`=\fracc3=144`

`|c|c^2=216`

Since
`216=6^3`, you have
`c=\pm6`.