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Question: OX and OY are two coordinates axes. On OY is taken a fixed point P and on OX any point Q. On PQ an e...

OX and OY are two coordinates axes. On OY is taken a fixed point P and on OX any point Q. On PQ an equilateral triangle is described, its vertex R being on the side of PQ away from O, then the locus of R will be –

A

Straight line

B

Circle

C

Ellipse

D

Parabola

Answer

Straight line

Explanation

Solution

Let P be (0, c), c is a constant number

From the figure, h = OL = OQ + QL

= c cot q + QR cos (180 – 60 – q)

= c cot q + PQ {cos 120 . cos q + sin 120 . sin q}

= c cot q – PQ2\frac{PQ}{2} cos q + 32\frac{\sqrt{3}}{2} PQ sin q

= c cot q – 12\frac{1}{2}c coses q cos q + 32\frac{\sqrt{3}}{2} c coses q sin q

[sinθ=cPQ]\left\lbrack \because\sin\theta = \frac{c}{PQ} \right\rbrack

= c2\frac{c}{2} (cot q + 3\sqrt{3}) … (i)

Again, k = RL = RQ sin (180 – 60 – q) = PQ

{sinθ12+cosθ32}\left\{ \sin\theta\frac{1}{2} + \cos\theta\frac{\sqrt{3}}{2} \right\}

= c cosec q {12sinθ+32cosθ}\left\{ \frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta \right\}

= c2\frac{c}{2} (1 + 3\sqrt{3} cot q) … (ii)

Eliminating cot q from (i) and (ii), we get

k = 3\sqrt{3}h – c

\The required locus is y =3\sqrt{3}x – c, a straight line.