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

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Question

Given Ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$

$e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2}$

$\because P$ is a point on the ellipse

So, $P

Vedclass · Accepted Answer

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

Vedclass · Answer

Given Ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$

$e=\sqrt{1-\frac{b^2}{a^2}}=\frac{\sqrt{3}}{2}$

$\because P$ is a point on the ellipse

So, $P =(4 \cos 	heta, 2 \sin 	heta)$

Equation of normal to the ellipse $\frac{x^2}{16}+\frac{y^2}{4}=1$ at point $\left( x _1, y _1ight)=(4 \cos 	heta, 2 \sin 	heta)$ is given by

$a ^2 y _1\left( x - x _1ight)= b ^2 x _1\left( y - y _1ight)$

$\Rightarrow 16 	imes 2 \sin 	heta( x -4 \cos 	heta)=4 	imes 4 \cos 	heta( y -2 \sin 	heta)$

$\Rightarrow 2 x \sin 	heta-8 \sin 	heta \cos 	heta= y \cos 	heta-2 \sin 	heta \cos 	heta)$

$\Rightarrow 2 x \sin 	heta= y \cos 	heta+6 \sin 	heta \cos 	heta$

$\Rightarrow \frac{2 x }{\cos 	heta}=\frac{ y }{\sin 	heta}+6$

$\Rightarrow 2 x \sec 	heta- y \operatorname{cosec} 	heta=6$

It meet the $x$-axis at $Q(3 \cos 	heta, 0)$

$	herefore M =\left(\frac{7}{2} \cos 	heta, \sin 	hetaight)=( x , y )$

Locus of $M$ is

$\frac{x^2}{\left(\frac{7}{2}ight)^2}+\frac{y^2}{1}=1$

Latus rectum of the given ellipse is

$x = \pm ae = \pm \sqrt{16-4}= \pm 2 \sqrt{3}$

So locus of $M$ meets the latus rectum at points for which

$y ^2=1-\frac{12 	imes 4}{49}=\frac{1}{49} \Rightarrow y = \pm \frac{1}{7}$

Hence, the required point is $\left( \pm 2 \sqrt{3}, \pm \frac{1}{7}ight)$.

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

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