Question

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$L^{2}$ = PM · PN, then locus of P is-

• A. x² + y² + ax –$\frac{a}{\sqrt{3}}$y = 0
• B. x² + y² – ax –$\frac{a}{\sqrt{3}}$y = 0
• C. x² + y² – ax +$\frac{a}{\sqrt{3}}$y = 0
• D. None of these

Answer 1

Answer: (C)

x² + y² – ax +$\frac{a}{\sqrt{3}}$y = 0

Answer 2

Let us choose A as the origin and AB as the X-axis. Then B ŗ (a, 0) and the equations of lines AC and BC, are respectively given by

y – $\sqrt{3}$x = 0

y + $\sqrt{3}$ (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 PL² = PM · PN

i. e. k² = $\frac{|k - \sqrt{3}h|}{2}$ · $\frac{|k + \sqrt{3}(h - a)|}{2}$

i.e. 4k² = (k – $\sqrt{3}$h) [k + $\sqrt{3}$ (h – a)] [Q P lies below both the lines]

i.e. 3(h² + k²) – 3ah +$\sqrt{3}$ak = 0

i.e. h² + k² – ah +$\frac{a}{\sqrt{3}}$k = 0

Putting (x, y) in place of (h, k) gives the equation of the required locus as

x² + y² – ax + $\frac{a}{\sqrt{3}}$y = 0.