[tex]x=y^2-1\implies y=\pm\sqrt{x+1}[/tex]
We want to find [tex]a[/tex] such that
[tex]\displaystyle\int_{-1}^a2\sqrt{x+1}\,\mathrm dx=\int_a^02\sqrt{x+1}\,\mathrm dx[/tex]
On the left,
[tex]\displaystyle\int_{-1}^a2\sqrt{x+1}\,\mathrm dx=4(x+1)^{3/2}\bigg|_{x=-1}^{x=a}=4(a+1)^{3/2}[/tex]
and on the right,
[tex]\displaystyle\int_a^02\sqrt{x+1}\,\mathrm dx=4(x+1)^{3/2}\bigg|_{x=a}^{x=0}=4-4(a+1)^{3/2}[/tex]
For the areas to be equal, we need
[tex]4(a+1)^{3/2}=4-4(a+1)^{3/2}\implies(a+1)^{3/2}=\dfrac12[/tex]
[tex]\implies a=\dfrac1{2^{2/3}}-1\approx-0.370[/tex]
