Answer
$\frac{3}{4}$
Work Step by Step
Step 1. Using the figure given in the Exercise, with the known coordinates of the intersection, we can get the area of the triangle as $A_1=\frac{1}{2}(2a)(a^2)=a^3$
Step 2. Using symmetry, the area of the parabolic region can be obtained as $A_2=2\int_0^a(a^2-x^2)dx=2a^2x|_0^a-\frac{2}{3}x^3|_0^a=\frac{4}{3}a^3$
Step 3. Thus the ratio can be found as $\lim_{a\to0}\frac{A_1}{A_2}=\frac{3}{4}$ (Note that $a$ is cancelled in the ratio).
