The normal at a point $P$ on the ellipse $x^{2} + 4y^{2} = 16$ meets the $x$-axis at $Q$. If $M$ is the mid point of the line segment $PQ$, then the locus of $M$ intersects the latus rectums of the given ellipse at the points

The normal at a point P on the ellipse x² + 4y² = 16 meets the x-axis at Q. If M is the mid point of the line segment PQ, then the locus of M intersects the latus rectums of the given ellipse at the points $\text{Given: ellipse } x^2 + 4y^2 = 16$ $\text{Normal at } P \text{ meets } x\text{-axis at } Q$ $\text{To find: locus of } M \ \cap \ \text{latus rectums}$ $M = \text{midpoint of } PQ$ $\frac{x^2}{16} + \frac{y^2}{4} = 1$ $a=4,\quad b=2$ $1.\ \text{normal at } P \Rightarrow Q$ $2.\ M = \text{midpoint of } PQ \Rightarrow \text{locus}$ $3.\ \text{cut locus with latus rectums}$ $\frac{a^2 x}{x_1} - \frac{b^2 y}{y_1} = a^2 - b^2$ $\frac{16x}{x_1} - \frac{4y}{y_1} = 12$ $y=0:\ \frac{16x}{x_1}=12 \Rightarrow x=\frac{3x_1}{4}$ $Q\left(\frac{3x_1}{4},\ 0\right)$ $M=\left(\frac{x_1 + 3x_1/4}{2},\ \frac{y_1}{2}\right)$ $=\left(\frac{7x_1}{8},\ \frac{y_1}{2}\right)$ $X=\frac{7x_1}{8},\ Y=\frac{y_1}{2}$ $\Rightarrow x_1=\frac{8X}{7},\ y_1=2Y$ $\frac{(8X/7)^2}{16} + \frac{(2Y)^2}{4} = 1$ $\frac{4X^2}{49} + Y^2 = 1$ $\frac{x^2}{49/4} + y^2 = 1$ $e=\sqrt{1-\frac{b^2}{a^2}} = \sqrt{1-\frac{4}{16}} = \frac{\sqrt3}{2}$ $x=\pm ae = \pm 2\sqrt3 \quad (\text{latus rectums})$ $x^2 = 12:\ \frac{48}{49} + y^2 = 1$ $y^2 = \frac{1}{49} \Rightarrow y = \pm \frac{1}{7}$ $\left(\pm 2\sqrt3,\ \pm \frac{1}{7}\right)$ $\Rightarrow \text{option (3)}$