The normal at a point P on the ellipse x^2+4y | vedclass

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(C) Given ellipse is $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Here $a^2 = 16$ and $b^2 = 4$.Let $P = (4 \cos 	heta, 2 \sin 	heta)$ be a point on the el

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

Vedclass · Answer

(C) Given ellipse is $\frac{x^2}{16} + \frac{y^2}{4} = 1$. Here $a^2 = 16$ and $b^2 = 4$.Let $P = (4 \cos 	heta, 2 \sin 	heta)$ be a point on the ellipse.The equation of the normal at $P$ is $\frac{ax}{\cos 	heta} - \frac{by}{\sin 	heta} = a^2 - b^2$.Substituting $a=4, b=2$: $\frac{4x}{\cos 	heta} - \frac{2y}{\sin 	heta} = 16 - 4 = 12$,which simplifies to $\frac{x}{\cos 	heta} - \frac{y}{2 \sin 	heta} = 3$.This normal meets the $x$-axis $(y=0)$ at $Q(3 \cos 	heta, 0)$.Let $M(x, y)$ be the midpoint of $PQ$. Then $x = \frac{4 \cos 	heta + 3 \cos 	heta}{2} = \frac{7}{2} \cos 	heta$ and $y = \frac{2 \sin 	heta + 0}{2} = \sin 	heta$.Thus,$\cos 	heta = \frac{2x}{7}$ and $\sin 	heta = y$. Using $\cos^2 	heta + \sin^2 	heta = 1$,the locus of $M$ is $\frac{4x^2}{49} + y^2 = 1$.The latus rectum of the original ellipse is $x = \pm ae = \pm \sqrt{a^2 - b^2} = \pm \sqrt{16 - 4} = \pm \sqrt{12} = \pm 2 \sqrt{3}$.Substituting $x^2 = 12$ into the locus equation: $\frac{4(12)}{49} + y^2 = 1 \Rightarrow y^2 = 1 - \frac{48}{49} = \frac{1}{49}$.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 rectums of the given ellipse at the points

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