Figure 2 shows the approximate shape of the vertical cross section of the entrance to a cave - AQA - A-Level Maths Pure - Question 13 - 2022 - Paper 1

First, substituting $y = 0$ and $x = 16$ into the equation gives:

$16^2 + 0^2 = a\sqrt{16 - 0}$

This simplifies to:

$256 = a\sqrt{16}$

Since $\sqrt{16} = 4$, we can rewrite the equation as:

$256 = a \cdot 4$