The diagram below shows the graph of , for , where . The line L is the tangent to the graph of at…

Given $f'(x) = \frac{2a^2 - 4x^2}{\sqrt{a^2 - x^2}}$. The slope of the tangent L at the origin $O(0,0)$ is $m = f'(0) = \frac{2a^2 - 0}{\sqrt{a^2 - 0}} = \frac{2a^2}{a} = 2a$. The equation of tangent line L is $y - 0 = 2a(x - 0) \Rightarrow y = 2ax$. The point $P(a, b)$ lies on line L, so $b = 2a(a) = 2a^2$, giving $P(a, 2a^2)$. Point $Q$ is $(a, 0)$, and $O$ is $(0, 0)$. The triangle $\Delta OPQ$ is bounded by the line $y = 2ax$, the vertical line $x = a$, and the x-axis from $x = 0$ to $x = a$. Using integration, the area is: $\text{Area} = \int_0^a 2ax dx = 2a \left[\frac{x^2}{2}\right]_0^a = a [a^2 - 0] = a^3 \text{ sq. units}$.