Let $C$ be a variable point in the first quadrant on line $x+y=1$, which intersects the axes at $A$ and $B$ respectively. Let $D$ and $E$ be the foot of perpendiculars on the axes from the point $C$, And $P(h,k)$ be a point on the line segment $DE$.
We need to find
1.the maximum possible value of product $hk$.
2.the area traced by point $P$.
I assumed a point $C(\cos^2\theta,\sin^2\theta)$ ,giving us $$ \frac{h}{\cos^2\theta} + \frac{k}{\sin^2\theta}=1 $$ from here we can apply AM-GM inequality , yielding $$hk\leq \sin^22\theta/16$$
Making the maximum required value to be $1/16$, This solves first part of the question.
But now because $P$ only lies on curve $xy=1/16$ at one point $(1/4,1/4)$, We cannot use this function ($y=x/16$) to find the area of the locus of $P$ ( what we need here is the area of the envelope of the line segment $DE$
So my question is how should we visualise and find the area traced by point P?
Using your parameterization for $C$, line segment $DE$ has the equation
$y = \sin^2(\theta) - \tan^2(\theta) x $
which can written as
$ f(x, y, \theta) = y + \tan^2(\theta) x - \sin^2(\theta) = 0 $
we want to find the envelope of all such lines segments, and for that we differentiate $f(x, y, \theta)$ with respect to $\theta$ and set that equal to zero.
$\dfrac{\partial f}{\partial \theta} = 2 \tan(\theta) \sec^2(\theta) x - 2 \sin(\theta) \cos(\theta) x = 0 $
Multiplying through by $\cos^3(\theta) / \sin(\theta) $ results in
$ x = \cos^4(\theta) $
Substituting this into $f(x, y, \theta) = 0 $ gives
$ y = \sin^4(\theta) $
Now, we have to eliminate $\theta$ and obtain a relation between $x$ and $y$ directly. We note that
$ x + y = \cos^4(\theta) + \sin^4(\theta) = (\cos^2 (\theta) + \sin^2 (\theta))^2 - 2 \cos^2(\theta) \sin^2(\theta) = 1 - 2 \sqrt{x y}$
Hence,
$2 \sqrt{xy} = 1 - x - y $
Squaring both sides,
$ 4 x y = 1 + x^2 + y^2 - 2 x - 2 y + 2 x y $
which simplifies to
$ x^2 + y^2 - 2 x y - 2 x - 2 y + 1 = 0 $
which can be written as
$ (x - y)^2 - 2 (x + y) + 1 = 0 $
Introducing the change of coordinates
$ y' = \dfrac{1}{\sqrt{2}} (x + y) $
$ x' = \dfrac{1}{\sqrt{2}} (x - y) $
Then
$ 2 x'^2 - 2 \sqrt{2} y' + 1 = 0 $
Which is the parabola
$y' = \dfrac{1}{2\sqrt{2}} ( 2 x'^2 + 1 ) $
By sketching this curve, we find that the range for $x'$ is $[-\dfrac{1}{\sqrt{2}}, \dfrac{1}{\sqrt{2}} ] $
The axis of the parabola is $x' = 0$ with corresponds to $ y = x$
The $x$ axis ( $ y = 0 ) $ corresponds to $ y' - x' = 0 $ i.e. $y' = x'$
Now the required area is
$ A = 2 \displaystyle \int_0^{\frac{1}{\sqrt{2}}} \left( \dfrac{1}{2\sqrt{2}} ( 2 x'^2 + 1 ) - x' \right) d x' $
which is straight forward to evaluate
$A = 2 \left[ \dfrac{\sqrt{2}}{3} u^3 + \dfrac{1}{\sqrt{2}} u -\dfrac{1}{2} u^2 \right]_0^{\frac{1}{\sqrt{2}}} $
$= 2 \left[ \dfrac{1}{12} + \dfrac{1}{4} - \dfrac{1}{4} \right] = \boxed{\dfrac{1}{6}} $