We are asked to find the area of the largest rectangle that can be inscribed in the ellipse given by the equation $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.

1. Problem Description

We are asked to find the area of the largest rectangle that can be inscribed in the ellipse given by the equation $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.

2. Solution Steps

Let the center of the ellipse be $(h,k)$. Due to symmetry, we can assume that one of the vertices of the rectangle lies in the first quadrant relative to the center of the ellipse. Let this vertex be $(x,y)$. Then the vertices of the rectangle are $(h \pm (x-h), k \pm (y-k))$.

The width of the rectangle is $2(x-h)$ and the height is $2(y-k)$. Therefore, the area of the rectangle is

$A = 4(x-h)(y-k)$.

Let $x' = x-h$ and $y' = y-k$. Then the equation of the ellipse becomes $\frac{x'^2}{a^2} + \frac{y'^2}{b^2} = 1$, and the area becomes $A = 4x'y'$. We want to maximize $A$.

From the equation of the ellipse, we have $\frac{y'^2}{b^2} = 1 - \frac{x'^2}{a^2}$, so $y'^2 = b^2(1 - \frac{x'^2}{a^2})$, and $y' = b\sqrt{1 - \frac{x'^2}{a^2}}$.

Substituting this into the area equation, we have $A = 4x'b\sqrt{1 - \frac{x'^2}{a^2}} = \frac{4b}{a}x'\sqrt{a^2 - x'^2}$.

Let $f(x') = x'\sqrt{a^2 - x'^2}$. We want to maximize $f(x')$. To do this, we can maximize $f(x')^2 = x'^2(a^2 - x'^2) = a^2x'^2 - x'^4$. Let $u = x'^2$. Then we want to maximize $g(u) = a^2u - u^2$. Taking the derivative with respect to $u$, we get $g'(u) = a^2 - 2u$. Setting this equal to 0, we get $u = \frac{a^2}{2}$. So $x'^2 = \frac{a^2}{2}$, which means $x' = \frac{a}{\sqrt{2}}$.

Now we find $y' = b\sqrt{1 - \frac{x'^2}{a^2}} = b\sqrt{1 - \frac{a^2/2}{a^2}} = b\sqrt{1 - \frac{1}{2}} = b\sqrt{\frac{1}{2}} = \frac{b}{\sqrt{2}}$.

Then the maximum area is $A = 4x'y' = 4\left(\frac{a}{\sqrt{2}}\right)\left(\frac{b}{\sqrt{2}}\right) = 4\frac{ab}{2} = 2ab$.

3. Final Answer

The area of the largest rectangle is $2ab$.