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A point moves so that the sum of the squ...

A point moves so that the sum of the squares of the perpendiculars let fall from it on the sides of an equilateral triangle is constant. Prove that its locus is a circle.

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Let vertices B and C of equilateral triangle be (-a,0) and (a,0)
`OA=OB tan 60^(@)=a sqrt(3)`
`:. A -= (0,asqrt(3))`
Let point P be (h,k).
Equations of AC and BC are `xsqrt(3)+y-asqrt(3)=0` and `xsqrt(3)-y+asqrt(3)=0`, respectively.
According to the question,
`PL^(2)+PM^(2)+PN^(2)=` constant (say `lambda)`
`:. k^(2)+((sqrt(3)h+k-sqrt(3)a)/(2))^(2)+((sqrt(3)h-k+sqrt(3)a)/(2))^(2)=lambda`
`implies 6h^(2)+6k^(2)-4sqrt(3)ak+6a^(2)-4 lambda=0`
Hence, the required locus is `6x^(2)+6y^(2)-4sqrt(3)ay+6a^(2)-4 lambda =0` which is a circle
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