1. Write the ellipse in standard form

Given ellipse: $x^2+4y^2=16$ Divide by $16$: $\frac{x^2}{16}+\frac{y^2}{4}=1$ So, $a^2=16,\quad b^2=4,\quad a=4,\quad b=2$

Also, $c^2=a^2-b^2=16-4=12 \implies c=2\sqrt{3}$ Hence the latus rectums are the vertical lines: $x=\pm c=\pm 2\sqrt{3}$

2. Take a general point on the ellipse

Let $P=(x_1,y_1)$ on the ellipse, so $x_1^2+4y_1^2=16$

Differentiate implicitly: $2x+8y\frac{dy}{dx}=0$ $\frac{dy}{dx}=-\frac{x}{4y}$ So slope of tangent at $P$ is $m_t=-\frac{x_1}{4y_1}$ Therefore slope of normal is $m_n=\frac{4y_1}{x_1}$

Equation of the normal at $P$: $y-y_1=\frac{4y_1}{x_1}(x-x_1)$

3. Find point $Q$ where the normal meets the $x$-axis

Since $Q$ lies on the $x$-axis, its $y$-coordinate is $0$. Put $y=0$ in the normal equation: $-y_1=\frac{4y_1}{x_1}(x-x_1)$ Assuming $y_1\neq 0$ (the formula will still lead to the locus correctly), divide by $y_1$: $-1=\frac{4}{x_1}(x-x_1)$ $-x_1=4x-4x_1$ $4x=3x_1$ $x=\frac{3x_1}{4}$ Thus, $Q=\left(\frac{3x_1}{4},0\right)$

4. Find midpoint $M$ of $PQ$

If $M=(h,k)$ then $h=\frac{x_1+\frac{3x_1}{4}}{2}=\frac{7x_1}{8},\qquad k=\frac{y_1+0}{2}=\frac{y_1}{2}$ So, $x_1=\frac{8h}{7},\qquad y_1=2k$

Substitute into ellipse equation: $x_1^2+4y_1^2=16$ $\left(\frac{8h}{7}\right)^2+4(2k)^2=16$ $\frac{64h^2}{49}+16k^2=16$ Divide by $16$: $\frac{4h^2}{49}+k^2=1$ Thus locus of $M$ is $\frac{4x^2}{49}+y^2=1$

5. Intersect this locus with the latus rectums of the given ellipse

The latus rectums are: $x=\pm 2\sqrt{3}$ Substitute in the locus: $\frac{4(2\sqrt{3})^2}{49}+y^2=1$ $\frac{4\cdot 12}{49}+y^2=1$ $\frac{48}{49}+y^2=1$ $y^2=\frac{1}{49}$ $y=\pm \frac{1}{7}$

Therefore the intersection points are $\left(\pm 2\sqrt{3},\,\pm \frac{1}{7}\right)$

6. Match with options

This corresponds to Option C: $\left(\pm 2\sqrt{3},\,\pm \frac{1}{7}\right)$