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From any point \mathrm{P} on the ellipse \mathrm{\mathrm{PN} } is drawn perpendicular to the major axis and produced at \mathrm{\mathrm{Q} } so that \mathrm{\mathrm{NQ} } equals to \mathrm{\mathrm{PS}, where\, \mathrm{S} }is a focus. The locus of \mathrm{ \mathrm{Q} } is the two straight lines \mathrm{ \mathrm{y} \pm \mathrm{ex}+\mathrm{a}=0 }

Option: 1

y \pm e x+a=0


Option: 2

y+e x \pm a=0


Option: 3

y-e x \pm a=0


Option: 4

y \pm x+a e=0


Answers (1)

best_answer

Let any point \mathrm{\mathrm{P} } on the ellipse
\mathrm{\begin{aligned} & \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \text { be }\left(x_1, y_1\right) \text {, then } \\ & N Q=P S=a+e x_1 \end{aligned} }

Let the coordinate of \mathrm{Q_{\text {be }}(x, y) }, then
\mathrm{\mathrm{x}=\mathrm{CN}=\mathrm{x}_1..........................(1) }
\mathrm{And \, \, y=N Q=-\left(a+e x_1\right)..........................(2) }
Putting the value of  \mathrm{x_1 } from (1) in (2),
The required locus is \mathrm{ y=-(a+e x) }
\mathrm{\mathrm{y}+\mathrm{a}+\mathrm{ex}=0 }
Taking \mathrm{S^{\prime} } instead of S the locus of Q will be \mathrm{y+a-e x=0 }

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Shailly goel

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