ABC is an equilateral triangle of side ‘a’. L, M and N are foot of the perpendiculars drawn from a point P to the sides AB, BC and CA respectively. If P lies inside the triangle and satisfies the condition P 
= PM · PN, then locus of P is-
y = 0
y = 0
y = 0
Text Solution
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Let us choose A as the origin and AB as the X-axis. Then B ≡ (a, 0) and the equations of lines AC and BC, are respectively given by
y – 
x = 0
y + 
(x – a) = 0
Let P (h, k) be the coordinates of the point whose locus is to be found. Now, according to the given condition, we have PL 2 = PM · PN
i. e. k 2 = 
· 
i.e. 4k 2 = (k – 
h) [k + 
(h – a)] [ P lies below both the lines]
i.e. 3(h 2 + k 2 ) – 3ah + 
ak = 0
i.e. h 2 + k 2 – ah + 
k = 0
Putting (x, y) in place of (h, k) gives the equation of the required locus as
x 2 + y 2 – ax + 
y = 0.
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