The equation of a curve is $x^2 - 2xy + 2y^2 = 20$.

1. Differentiate the given equation of the curve implicitly with respect to $x$.

Differentiating $x^2 - 2xy + 2y^2 = 20$ with respect to $x$:
$\frac{d}{dx}(x^2) - \frac{d}{dx}(2xy) + \frac{d}{dx}(2y^2) = \frac{d}{dx}(20)$
Applying the product rule to the $xy$ term and the chain rule to the $y^2$ term:
$2x - \left( 2y + 2x \frac{dy}{dx} \right) + 4y \frac{dy}{dx} = 0$
$2x - 2y - 2x \frac{dy}{dx} + 4y \frac{dy}{dx} = 0$

2. Substitute the coordinates of point $P$ to find the gradient of the tangent.

At the point $P(2,4)$, substitute $x=2$ and $y=4$ into the differentiated equation:
$2(2) - 2(4) - 2(2)\frac{dy}{dx} + 4(4)\frac{dy}{dx} = 0$
$4 - 8 - 4\frac{dy}{dx} + 16\frac{dy}{dx} = 0$
$-4 + 12\frac{dy}{dx} = 0$
$12\frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{1}{3}$
The gradient of the tangent at $P$ is $\frac{1}{3}$.

3. Formulate the equation of the tangent at $P$.

Using the point-slope form $y - y_1 = m(x - x_1)$ with $m = \frac{1}{3}$ and $P(2,4)$:
$y - 4 = \frac{1}{3}(x - 2)$
Multiplying throughout by $3$:
$3y - 12 = x - 2$
$x - 3y + 10 = 0$

4. Determine the gradient of the normal.

Let the gradient of the normal be $m_n$. Since the normal is perpendicular to the tangent ($m_n \times m_t = -1$):
$m_n = -\frac{1}{1/3} = -3$

5. Formulate the equation of the normal at $P$.

Using the point-slope form with $m_n = -3$ and $P(2,4)$:
$y - 4 = -3(x - 2)$
$y - 4 = -3x + 6$
$3x + y - 10 = 0$

1. Find the coordinates of $A$ and $B$ from the equations of the tangent and normal.

From part (i), the equation of the tangent at $P(2,4)$ is $y = \frac{1}{3}x + \frac{10}{3}$ and the normal is $y = -3x + 10$.

To find $A$ (where the tangent meets the $y$-axis), substitute $x = 0$:
$y = \frac{10}{3} \implies A\left(0, \frac{10}{3}\right)$

To find $B$ (where the normal meets the $x$-axis), substitute $y = 0$:
$0 = -3x + 10 \implies x = \frac{10}{3} \implies B\left(\frac{10}{3}, 0\right)$

2. Identify the right angle and calculate the lengths of $PA$ and $PB$.

Since the tangent and normal are mutually perpendicular, $\triangle APB$ is a right-angled triangle at $P$ (i.e. $\angle APB = 90^\circ$).

Using the distance formula for the legs $PA$ and $PB$:
$PA = \sqrt{(2-0)^2 + \left(4-\frac{10}{3}\right)^2} = \sqrt{4 + \frac{4}{9}} = \sqrt{\frac{40}{9}} = \frac{2\sqrt{10}}{3}$
$PB = \sqrt{\left(2-\frac{10}{3}\right)^2 + (4-0)^2} = \sqrt{\frac{16}{9} + 16} = \sqrt{\frac{160}{9}} = \frac{4\sqrt{10}}{3}$

3. Compute the area of triangle $APB$.

Using the formula for the area of a right-angled triangle:
$\text{Area of } \triangle APB = \frac{1}{2} \times PA \times PB$
$\text{Area} = \frac{1}{2} \left(\frac{2\sqrt{10}}{3}\right) \left(\frac{4\sqrt{10}}{3}\right)$
$\text{Area} = \frac{1}{2} \times \frac{80}{9} = \frac{40}{9} \text{ units}^2$