Question:

Step 1: Determine the parameters of the ellipse. The given ellipse is $x^2 + 4y^2 = 16$. Dividing by 16, we get the standard form: $$ \frac{x^2}{16} + \frac{y^2}{4} = 1 $$ Comparing this with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, we have $a^2 = 16$ and $b^2 = 4$. Thus, $a = 4$ and $b = 2$. The distance from the center to the foci is $c$, where $c^2 = a^2 - b^2 = 16 - 4 = 12$. So, $c = \sqrt{12} = 2\sqrt{3}$. The eccentricity is $e = \frac{c}{a} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2}$. A general point P on the ellipse can be parameterized as $P(a\cos\theta, b\sin\theta)$, which is $P(4\cos\theta, 2\sin\theta)$. Step 2: Find the equation of the normal at point P. The equation of the normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at a point $(x_0, y_0)$ is given by: $$ \frac{a^2x}{x_0} - \frac{b^2y}{y_0} = a^2 - b^2 $$ Substituting $a^2=16$, $b^2=4$, $x_0=4\cos\theta$, and $y_0=2\sin\theta$: $$ \frac{16x}{4\cos\theta} - \frac{4y}{2\sin\theta} = 16 - 4 $$ $$ \frac{4x}{\cos\theta} - \frac{2y}{\sin\theta} = 12 $$ Step 3: Find the coordinates of point Q. Point Q is where the normal meets the x-axis. Setting $y=0$ in the normal equation: $$ \frac{4x}{\cos\theta} = 12 $$ $$ x = \frac{12\cos\theta}{4} = 3\cos\theta $$ So, the coordinates of Q are $(3\cos\theta, 0)$. Step 4: Find the midpoint M of PQ. P is $(4\cos\theta, 2\sin\theta)$ and Q is $(3\cos\theta, 0)$. The midpoint M is given by: $$ M = \left(\frac{4\cos\theta + 3\cos\theta}{2}, \frac{2\sin\theta + 0}{2}\right) $$ $$ M = \left(\frac{7\cos\theta}{2}, \sin\theta\right) $$ Step 5: Determine the locus of M. Let the coordinates of M be $(h, k)$. From Step 4, we have: $$ h = \frac{7\cos\theta}{2} \implies \cos\theta = \frac{2h}{7} $$ $$ k = \sin\theta $$ Using the trigonometric identity $\cos^2\theta + \sin^2\theta = 1$: $$ \left(\frac{2h}{7}\right)^2 + k^2 = 1 $$ $$ \frac{4h^2}{49} + k^2 = 1 $$ Multiplying by 49, we get: $$ 4h^2 + 49k^2 = 49 $$ Replacing $(h, k)$ with $(x, y)$, the locus of M is: $$ 4x^2 + 49y^2 = 49 $$ Step 6: Determine the equations of the latus rectums of the given ellipse. The foci of the ellipse $x^2 + 4y^2 = 16$ are at $(\pm c, 0)$, which are $(\pm 2\sqrt{3}, 0)$. The latus rectums are vertical lines passing through the foci. Their equations are: $$ x = \pm 2\sqrt{3} $$ Step 7: Find the intersection points of the locus of M and the latus rectums. Substitute $x = \pm 2\sqrt{3}$ into the locus equation $4x^2 + 49y^2 = 49$: $$ 4(\pm 2\sqrt{3})^2 + 49y^2 = 49 $$ $$ 4(4 \times 3) + 49y^2 = 49 $$ $$ 4(12) + 49y^2 = 49 $$ $$ 48 + 49y^2 = 49 $$ $$ 49y^2 = 1 $$ $$ y^2 = \frac{1}{49} $$ $$ y = \pm \frac{1}{7} $$ Therefore, the locus of M intersects the latus rectums of the given ellipse at the points $\left(\pm 2\sqrt{3}, \pm \frac{1}{7}\right)$.