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
\(32x^2=2c^2\implies 16x^2=c^2\implies x=\pm\dfrac{|c|}4\)
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
\(\displaystyle2\int_0^{|c|/4}(2c^2-32x^2)\,\mathrm dx=4\int_0^{|c|/4}(c^2-16x^2)\,\mathrm dx\)
\(=4\left[c^2x-\dfrac{16}3x^3\right]_{x=0}^{x=|c|/4}\)
\(=c^2|c|-\dfrac{|c|^3}3\)
\(=\dfrac{2|c|c^2}3=144\)
\(|c|c^2=216\)
Since \(216=6^3\), you have \(c=\pm6\).
