Equation of the locus of the centroid of the triangle whose vertices are (a cos k, a sin k) (b sin k,-b cos k) and (1,0), where k is a parameter, is

Question

Equation of the locus of the centroid of the triangle whose vertices are (a cos k, a sin k) (b sin k,-b cos k) and (1,0), where k is a parameter, is (A) (1-3 x)2+9 y2=a2+b2 (B) (3 x-1)2+9 y2=2 a2+2 b2 (C) (3 x+1)2+(3 y)2=a2+b2 (D) (3 x+1)2+(3 y)2=3 a2+3 b2

Tardigrade Admin · Accepted Answer

We have, A =(a cos k, a sin k) B =(b sin k,-b cos k) C =(1,0) Let (x, y) be the centroid of Δ A B C. ∴ x=(a cos k+b sin k+1/3) and y=(a sin k-b cos k+0/3) ⇒ 3 x-1=a cos k+b sin k ...(i) and 3 y=a sin k-b cos k ...(ii) On squaring and then adding Eqs. (i) and (ii) we get (3 x-1)2+(3 y)2=a2(. sin 2 k+ cos 2 k) +b2( sin 2 k+ cos 2 k) ∴ (3 x-1)2+9 y2=a2+b2 ∴ (1-3 x)2+9 y2=a2+b2

Q. Equation of the locus of the centroid of the triangle whose vertices are $(a cos k, a sin k)$ $(b sin k, - b cos k)$ and $(1, 0)$ , where $k$ is a parameter, is

$(1 - 3 x)^{2} + 9 y^{2} = a^{2} + b^{2}$

$(3 x - 1)^{2} + 9 y^{2} = 2 a^{2} + 2 b^{2}$

$(3 x + 1)^{2} + (3 y)^{2} = a^{2} + b^{2}$

$(3 x + 1)^{2} + (3 y)^{2} = 3 a^{2} + 3 b^{2}$

Q. Equation of the locus of the centroid of the triangle whose vertices are (acosk,asink) (bsink,−bcosk) and (1,0), where k is a parameter, is

(1−3x)2+9y2=a2+b2

(3x−1)2+9y2=2a2+2b2

(3x+1)2+(3y)2=a2+b2

(3x+1)2+(3y)2=3a2+3b2

Q. Equation of the locus of the centroid of the triangle whose vertices are $(a cos k, a sin k)$ $(b sin k, - b cos k)$ and $(1, 0)$ , where $k$ is a parameter, is

$(1 - 3 x)^{2} + 9 y^{2} = a^{2} + b^{2}$

$(3 x - 1)^{2} + 9 y^{2} = 2 a^{2} + 2 b^{2}$

$(3 x + 1)^{2} + (3 y)^{2} = a^{2} + b^{2}$

$(3 x + 1)^{2} + (3 y)^{2} = 3 a^{2} + 3 b^{2}$