For there to be a region bounded by the two parabolas, you first need to find some conditions on [tex]c[/tex]. The two parabolas must intersect each other twice, so you need two solutions to
[tex]16x^2-c^2=c^2-16x^2[/tex]
You have
[tex]32x^2=2c^2\implies 16x^2=c^2\implies x=\pm\dfrac{|c|}4[/tex]
which means you only need to require that [tex]c\neq0[/tex]. With that, the area of any such bounded region would be given by the integral
[tex]\displaystyle\int_{-4|c|}^{4|c|}\bigg((c^2-16x^2)-(16x^2-c^2)\bigg)\,\mathrm dx[/tex]
since [tex]c^2-16x^2>16x^2-c^2[/tex] for all [tex]c\neq0[/tex]. Now,
[tex]\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[/tex]
by symmetry across the y-axis. Integrating yields
[tex]\displaystyle2\int_0^{|c|/4}(2c^2-32x^2)\,\mathrm dx=4\int_0^{|c|/4}(c^2-16x^2)\,\mathrm dx[/tex]
[tex]=4\left[c^2x-\dfrac{16}3x^3\right]_{x=0}^{x=|c|/4}[/tex]
[tex]=c^2|c|-\dfrac{|c|^3}3[/tex]
[tex]=\dfrac{2|c|c^2}3=144[/tex]
[tex]|c|c^2=216[/tex]
Since [tex]216=6^3[/tex], you have [tex]c=\pm6[/tex].
