The ellipse has equation $$\frac{x^2}{16}+\frac{y^2}{9}=1$$ Applying $\frac{d}{dx}$ you get
\begin{align}
\frac x8 + \frac{2y}{9} \frac{dy}{dx}&=0 \\
\implies \frac{dy}{dx} & = -\frac{9x}{16y}
\end{align}
i.e. at a point $(x,y)$ on the ellipse, the tangent has gradient $-\frac{9x}{16y}$.
Parameterize the ellipse as $(x(t),y(t))=(4cost,3sint)$ where $0≤t<2 \pi$. Then at such a point, the tangent has gradient $-\frac{9(4cost)}{16(3sint)}=-\frac 34 cot(t)$, so that the equation of the tangent is $$y=-\frac 34 x cot(t)+3csc(t)$$
This line intersects the $x$ and $y$ axes at the points $(4sec(t),0)$ and $(0,3csc(t))$respectively. The midpoint between between these two points would be $(2sec(t),1.5csc(t))$, and recall that $0≤t<2\pi$.
So, in parametric form, the locus of midpoints of tangents is
\begin{align}
& x=\frac{2}{cos(t)} \\
& y=\frac{3}{2sin(t)} \\
\end{align}
i.e. $\frac 1x = \frac 12 cos(t)$ and $\frac 1y = \frac 23 sin(t)$, and we observe that they satisfy $4 \frac {1}{x^2} + \frac 94 \frac {1}{y^2} = 1$, hence $$\frac{16}{x^2}+\frac{9}{y^2}=4$$ as required.
If the end points of one axis are $A, B$, and the other semi-axis has length $w$, then the center of the ellipse is
$ C = \dfrac{1}{2} ( A_x + B_x , A_y + B_y ) $
The first semi-axis vector is
$ V_1 = \dfrac{1}{2} ( A_x - B_x, A_y - B_y ) $
And the second semi-axis vector is
$ V_2 = \dfrac{w}{ \| V_1 \| } \dfrac{1}{2} ( B_y - A_y, A_x - B_x ) $
Now the parametric equation of the ellipse is
$ p(t) = C + V_1 \cos t + V_2 \sin t $
Best Answer
With the given parametric expressions we have the derivative of the coordinates with respect to the parameter as $$\left\{ \begin{aligned} x &= a \cos\theta \\ y &= b \sin\theta \end{aligned} \right. \quad \implies \left\{ \begin{aligned} \dot{x} &= -a \sin\theta \\ \dot{y} &= b \cos\theta \end{aligned} \right. \qquad \text{where}~~ \dot{x} \equiv \frac{dx}{d\theta},~\dot{y} \equiv \frac{dy}{d\theta}~.$$ Thus the slope of the tangent line at point $P = (a\cos\theta, b\sin\theta)$ is $$\frac{dy}{dx} = \frac{ \dot{y} }{ \dot{x} } = -\frac{ b\cos\theta}{a\sin\theta}~,$$ and the slope of the normal line is the negative inverse of that: $$m \equiv -\left( \frac{dy}{dx} \right)^{-1} = \frac{ -\dot{x} }{ \dot{y} } = \frac{ a\sin\theta}{ b\cos\theta}~.$$ Note that here the parameter happens to be the polar angle, therefore $m$ is just the $\tan\theta$ (the slope of a normal line to a circle at $\theta$ angle) scaled by $\frac{a}b$.
The equation to the normal line is thus $$ \frac{ y - b\sin\theta }{ x - a\cos\theta } = m = \frac{ a\sin\theta}{ b\cos\theta}~,$$ or equivalently $$y - b \sin\theta = \frac{ a\sin\theta}{ b\cos\theta} (x - a\cos\theta)~ \tag*{Eq.(1)} $$ or to compare it with what is stated in the question post: $$\frac{ y \cos\theta }a - \frac{ x \sin\theta }b = \frac{ b^2 - a^2}{ ab} \sin\theta \cos\theta ~.$$
Denote $(x_0, y_0)$ as the coordinates of point $N$, the intersection of the normal line with the major axis (which is also the $x$-axis). The $x$-coordinate is readily calculated with $\text{Eq.}\,\text{(1)}$ at $y_0 = 0$
\begin{align} && y_0 - b \sin\theta &= \frac{ a\sin\theta}{ b\cos\theta} (x_0 - a\cos\theta) \\ &\implies & \frac{ - b^2 }a \cos\theta &= x_0 - a\cos\theta \\ &\implies & x_0 &= \left( a - \frac{ b^2 }a \right) \cos\theta \end{align} Denoting $r \equiv a - \frac{ b^2 }a = \frac{ a^2 - b^2 }a $, we have the center of the line segment $\overline{PN}$ being at $$ \left( \frac{ x + x_0}2 ~, \frac{ y + y_0}2\right) = \left( \frac{ a + r}2 \cos\theta ~, \frac{ b}2 \sin\theta\right) = \left( \frac{ 2a^2 - b^2}{ 2 a } \cos\theta ~, \frac{ b}2 \sin\theta\right) $$ At this point, we can identify the requested locus as an ellipse centere at the same origin and parametrized (necessarily) by the same parameter $\theta$ with the principle axes $$a' \equiv \frac{ 2a^2 - b^2}{ 2 a} = a - \frac{b^2}{ 2a } ~~, ~~b' \equiv \frac{b}2$$
Further Discussion
Note that the ratio $\frac{ a' }{ b'} \neq \frac{a}b$. That is, the new ellipse is NOT similar to the original. Nor does the ratio equal to $\sqrt{ \frac{a}b }$ or $\frac{a^2}{ b^2}$ or anything particularly nice. What we can say about the shape is that this new ellipse (from the locus) is more "elongated" than the original. Consider the wide-and-short original ellipse where $a > b$, we have $$ \frac{ a' }{ b'} = \frac{2a}b - \frac{b}a = \frac{a}b + \left( \frac{a}b - \frac{b}a \right) > \frac{a}b $$ where the larger ratio gives a more stretched out ellipse.
Also note that the locus goes "outside" of the original two foci, $$a' > \sqrt{a^2 - b^2} \qquad \because (a')^2 = a^2 - b^2 + \frac{ b^4 }{ 4a^2 } > a^2 - b^2$$ with its own foci that are closer to the center. $$(a')^2 - (b')^2 = a^2 - b^2 + \frac{ b^2}4 \left( \frac{ b^2 }{ a^2 } - 1\right) < a^2 - b^2$$