The normal at a point P on the ellipse x^2 + | vedclass

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(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Here $a^2 = 16$ and $b^2 = 4$. Let $P = (4 \cos 	heta, 2 \sin 	heta)$. The

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$\left( \pm 2\sqrt{3}, \pm \frac{1}{7} \right)$

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(C) The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Here $a^2 = 16$ and $b^2 = 4$. Let $P = (4 \cos 	heta, 2 \sin 	heta)$. The equation of the normal at $P$ is $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$. Substituting $a=4, b=2$,we get $\frac{4x}{\cos 	heta} - \frac{2y}{\sin 	heta} = 16 - 4 = 12$,which simplifies to $\frac{2x}{\cos 	heta} - \frac{y}{\sin 	heta} = 6$. For point $Q$ on the $x$-axis,$y=0$,so $x_Q = 3 \cos 	heta$. Thus $Q = (3 \cos 	heta, 0)$. Let $M = (h, k)$ be the midpoint of $PQ$. $h = \frac{4 \cos 	heta + 3 \cos 	heta}{2} = \frac{7}{2} \cos 	heta \implies \cos 	heta = \frac{2h}{7}$. $k = \frac{2 \sin 	heta + 0}{2} = \sin 	heta \implies \sin 	heta = k$. Using $\cos^2 	heta + \sin^2 	heta = 1$,we get $\frac{4h^2}{49} + k^2 = 1$. The locus of $M$ is $\frac{4x^2}{49} + y^2 = 1$. The latus rectum of the ellipse $\frac{x^2}{16} + \frac{y^2}{4} = 1$ is at $x = \pm ae$. $e = \sqrt{1 - \frac{b^2}{a^2}} = \sqrt{1 - \frac{4}{16}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}$. $ae = 4 	imes \frac{\sqrt{3}}{2} = 2\sqrt{3}$. Substitute $x = \pm 2\sqrt{3}$ into the locus equation: $\frac{4(12)}{49} + y^2 = 1 \implies y^2 = 1 - \frac{48}{49} = \frac{1}{49}$. So $y = \pm \frac{1}{7}$. The points are $\left( \pm 2\sqrt{3}, \pm \frac{1}{7} ight)$.

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

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