2021 AMC 10A Problem 24

Note that each of the equations yields two parallel lines. $$(x + ay)^2 = 4a^2$$ results in the two lines $$x + ay - 2a = 0$$ and $$x + ay + 2a = 0.$$ Both of these lines have a slope of $-\dfrac{1}{a}.$ Similarly, $$(ax-y)^2 = a^2$$ results in the lines $$ax - y - a = 0$$ and $$ax - y + a = 0.$$ These lines have slope $a.$ Note that each pair of lines is perpendicular to the other pair of lines. This shows that the equations form a rectangle. Recall that the formula for the distance $d$ between two parallel lines $$\begin{cases} Ax+By+C_1=0 \\ Ax+By+C_2=0 \end{cases}$$ is $$d = \dfrac{\mid C_2 - C_1 \mid}{\sqrt{A^2 + B^2}}.$$ Using this formula, we get that the distance between the first pair of lines is $$\dfrac{4a}{\sqrt{a^2 + 1}}.$$ Similarly, the distance between the second pair of lines is $$\dfrac{2a}{\sqrt{a^2 + 1}}.$$ These are the side lengths of the rectangle. Multiplying yields the area $$\dfrac{8a^2}{a^2 + 1}.$$ Thus, D is the correct answer.