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  • The normal at a point P on the ellipse meets the x-axis at Q.

The normal at a point P on the ellipse \mathrm{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 rectum of the given ellipse at the points

Option: 1

\mathrm{\left( \pm \frac{3 \sqrt{5}}{2}, \pm \frac{2}{7}\right)}


Option: 2

\mathrm{\left( \pm \frac{3 \sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4}\right)}


Option: 3

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


Option: 4

\mathrm{\left( \pm 2 \sqrt{3}, \pm \frac{4 \sqrt{3}}{7}\right)}


Answers (1)

best_answer

\mathrm{\text { The normal to the ellipse } \frac{x^2}{16}+\frac{y^2}{4}=1}

\mathrm{\text { at } P(4 \cos \theta, 2 \sin \theta) \text { is }}

\mathrm{4 x \sec \theta-2 y \operatorname{cosec} \theta=12 \text { which gives } Q=(3 \cos \theta, 0)}

\mathrm{\text { Let } M=(\alpha, \beta) \text { then } \alpha=\frac{3 \cos \theta+4 \cos \theta}{2}=\frac{7}{2} \cos \theta}

\mathrm{\beta=\sin \theta}

Eliminating ‘θ’ between the two equations

\mathrm{\left(\frac{2 \alpha}{7}\right)^2+(\beta)^2=1 \Rightarrow \frac{4 \alpha^2}{49}+\beta^2=1 \text { i.e., } \frac{4 x^2}{49}+y^2=1}

is the locus of M.
The latus rectum of the given ellipse is
x = ± 2 3
The locus of M meets the latus rectum whose y-coordinate is

\mathrm{\frac{48}{49}+y^2=1 \Rightarrow y= \pm \frac{1}{7}}

\mathrm{\text { Thus the points are }\left( \pm 2 \sqrt{3}, \pm \frac{1}{7}\right) \text {. }}

Posted by

Ritika Jonwal

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