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\mathrm{\mathrm{OX}\, and\, \mathrm{OY}} are two straight lines at right angles to one another; on \mathrm{\mathrm{OY} }is taken a fixed point A and on O X any variable point B. On A B an equilateral triangle is described, its vertex P being on the opposite side of A B to that of O. The locus of P is:

Option: 1

Straight line
 


Option: 2

 Circle
 


Option: 3

Ellipse
 


Option: 4

Parabola


Answers (1)

best_answer

Let the point  \mathrm{A} be  \mathrm{{ }^{(0, a)} } and \mathrm{\mathrm{B} } be \mathrm{(\mathrm{b}, 0) }
Let \mathrm{\mathrm{P} } be the moving point as (x, y) and let \mathrm{ \angle A B O=\theta } and P N perpendicular to OX

\mathrm{\begin{gathered} \mathrm{AB}=\mathrm{a} \operatorname{cosec} \theta=\mathrm{BP} \\ \mathrm{x}=\mathrm{ON}=\mathrm{OB}+\mathrm{BN}=\mathrm{OB}+\mathrm{BP} \cos \angle \mathrm{PBN} \\ \mathrm{x}=\mathrm{AB} \cos \theta+\mathrm{AB} \cos \left(120^{\circ}-\theta\right) \end{gathered} }

\mathrm{\begin{aligned} & \mathrm{x}=\mathrm{a} \operatorname{cosec} \theta \cdot \cos \theta+\mathrm{a} \operatorname{cosec} \theta\left(\cos 120^{\circ} \cdot \cos \theta+\sin 120^{\circ} \cdot \sin \theta\right) \\ & \mathrm{x}=\mathrm{a} \cot \theta-\frac{1}{2} \mathrm{a} \cot \theta+\mathrm{a} \frac{\sqrt{3}}{2} \\ & \mathrm{x}=\frac{\mathrm{a}}{2}(\cot \theta+\sqrt{3})...............(1) \\ & \text { Again } \mathrm{y}=\mathrm{PN}=\mathrm{PB} \sin \left(120^{\circ}-\theta\right) \\ & =\mathrm{a} \operatorname{cosec} \theta\left(\sin 120^{\circ} \cdot \cos \theta-\cos 120^{\circ} \cdot \sin \theta\right) \\ & \mathrm{y}=\mathrm{a} \cot \theta \cdot \frac{\sqrt{3}}{2}+\frac{\mathrm{a}}{2}=\frac{\mathrm{a}}{2}(\sqrt{3} \cot \theta+1).........................(2) \end{aligned} }
Eliminating \mathrm{\theta } from (1) and (2), we get  \mathrm{\sqrt{3} x-y=a } which represent a straight line.

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