2021 AMC 10A Problem 24

1. First see what the graphs are. If $(x + ay)^2 = 4a^2$, then $x + ay$ is a number whose square is $4a^2$, so $x + ay = 2a$ or $x + ay = -2a$. That is two parallel lines. In the same way, $(ax - y)^2 = a^2$ gives the two parallel lines $ax - y = a$ and $ax - y = -a$.
2. The answer has to be right for every $a$, so we can choose a value of $a$, find the area, and see which choice matches. Don't use $a = 1$: choices (C), (D) and (E) all equal $4$ there. Use $a = 2$.
3. With $a = 2$ the four lines are
$$x + 2y = 4, \quad x + 2y = -4, \quad 2x - y = 2, \quad 2x - y = -2.$$
4. Find the corners by solving pairs of equations. For example, take $x + 2y = 4$ and $2x - y = -2$. The second says $y = 2x + 2$. Put that into the first: $x + 4x + 4 = 4$, so $x = 0$ and $y = 2$. That corner is $(0, 2)$.
Doing the same for the other pairs gives the corners $(0, 2)$, $(\frac85, \frac65)$, $(0, -2)$ and $(-\frac85, -\frac65)$.
5. Two corners, $(0, 2)$ and $(0, -2)$, lie on the $y$-axis, $4$ apart. The segment joining them cuts the shape into two triangles. Each has base $4$ along the $y$-axis and height $\frac85$ (how far the other corner is from the $y$-axis). So the area is
$$2 \times \tfrac12 \times 4 \times \tfrac85 = \tfrac{32}{5}.$$
6. Put $a = 2$ into each choice:

Only (D) gives $\frac{32}{5}$.

The answer is $\boxed{\textbf{(D)}\ \frac{8a^2}{a^2+1}}$.

"Something squared equals a number" means the something is plus or minus the square root, so each equation hides two parallel lines. When a question asks for a formula that works for every value, testing one value against the choices is a fair and fast method, as long as the value you pick separates the choices.

Work with general $a$. The first pair of lines has slope $-\frac1a$ and the second has slope $a$. These multiply to $-1$, so the lines meet at right angles and the shape is a rectangle. Its sides are the gaps between each pair of parallel lines. The gap between the lines $Ax + By = c_1$ and $Ax + By = c_2$ is $\frac{|c_1 - c_2|}{\sqrt{A^2 + B^2}}$, which gives
$$\frac{4a}{\sqrt{a^2 + 1}} \quad\text{and}\quad \frac{2a}{\sqrt{a^2 + 1}}.$$
Their product is $\frac{8a^2}{a^2 + 1}$.

Take square roots: the first graph is the pair of parallel lines $x + ay = 2a$ and $x + ay = -2a$; the second is the pair $ax - y = a$ and $ax - y = -a$.

The first pair has normal vector $(1, a)$ and the second has normal vector $(a, -1)$. Their dot product is $a - a = 0$, so the two families are perpendicular, and the quadrilateral they bound is a rectangle.

The side lengths of the rectangle are the distances between the parallel lines in each pair, using $\frac{|c_1 - c_2|}{\sqrt{A^2 + B^2}}$:
$$\frac{2a - (-2a)}{\sqrt{1 + a^2}} = \frac{4a}{\sqrt{a^2+1}},\qquad \frac{a - (-a)}{\sqrt{a^2 + 1}} = \frac{2a}{\sqrt{a^2+1}}.$$

The area is their product:
$$\frac{4a}{\sqrt{a^2+1}} \cdot \frac{2a}{\sqrt{a^2+1}} = \frac{8a^2}{a^2+1}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{8a^2}{a^2+1}}$.