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 at $(h, k)$.

We can make a change of coordinates $x' = x - h$ and $y' = y - k$, so the equation of the ellipse becomes $\frac{x'^2}{a^2} + \frac{y'^2}{b^2} = 1$.

Consider a rectangle inscribed in the ellipse, with vertices $(x', y')$, $(-x', y')$, $(-x', -y')$, and $(x', -y')$. The area of the rectangle is $A = (2x')(2y') = 4x'y'$.

We want to maximize $A$ subject to the constraint $\frac{x'^2}{a^2} + \frac{y'^2}{b^2} = 1$.

From the ellipse equation, 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 $y' = b\sqrt{1 - \frac{x'^2}{a^2}}$.

Then the area $A = 4x'y' = 4x'b\sqrt{1 - \frac{x'^2}{a^2}} = \frac{4b}{a}x'\sqrt{a^2 - x'^2}$.

Let $f(x') = x'\sqrt{a^2 - x'^2}$. Then $f'(x') = \sqrt{a^2 - x'^2} + x'\frac{-2x'}{2\sqrt{a^2 - x'^2}} = \sqrt{a^2 - x'^2} - \frac{x'^2}{\sqrt{a^2 - x'^2}} = \frac{a^2 - x'^2 - x'^2}{\sqrt{a^2 - x'^2}} = \frac{a^2 - 2x'^2}{\sqrt{a^2 - x'^2}}$.

To find the critical points, we set $f'(x') = 0$, which gives $a^2 - 2x'^2 = 0$, so $2x'^2 = a^2$, and $x' = \frac{a}{\sqrt{2}}$.

Then $y' = b\sqrt{1 - \frac{x'^2}{a^2}} = b\sqrt{1 - \frac{a^2}{2a^2}} = b\sqrt{1 - \frac{1}{2}} = b\sqrt{\frac{1}{2}} = \frac{b}{\sqrt{2}}$.

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

3. Final Answer

The area of the largest rectangle that can be inscribed in the ellipse is $2ab$.