Question
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 –
Straight line
Circle
Ellipse
Parabola
Straight line
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 – 2PQ cos q + 23 PQ sin q
= c cot q – 21c coses q cos q + 23 c coses q sin q
[∵sinθ=PQc]
= 2c (cot q + 3) … (i)
Again, k = RL = RQ sin (180 – 60 – q) = PQ
{sinθ21+cosθ23}
= c cosec q {21sinθ+23cosθ}
= 2c (1 + 3 cot q) … (ii)
Eliminating cot q from (i) and (ii), we get
k = 3h – c
\The required locus is y =3x – c, a straight line.