Locus of the centroid of a triangle whose vertices are (1,0),(a cos t, a sin t),(b sin t,-b cos t) is 9 x2+9 y2-6 x=k. Then the value of k=

Question

Locus of the centroid of a triangle whose vertices are (1,0),(a cos t, a sin t),(b sin t,-b cos t) is 9 x2+9 y2-6 x=k. Then the value of k= (A) a2+b2 (B) a2+b2-1 (C) a2+b2+1 (D) 0

Tardigrade Admin · Accepted Answer

Centroid is given by x=(x1+x2+x3/3) and y=(y1+y2+y3/3) So, x=(1+a cos t+b sin t/3) and y=(0+a sin t+(-b cos t)/3) ⇒ (3 x-1)=a cos t+b sin t ldots (i) and 3 y=a sin t-b cos t ldots (2) Squaring and adding Eqs. (i) and (ii), we get 9 x2-6 x+1+9 y2=a2+b2 ⇒ 9 x2+9 y2-6 x=a2+b2-1 ∴ k=a2+b2-1

Q. Locus of the centroid of a triangle whose vertices are $(1, 0), (a cos t, a sin t), (b sin t, - b cos t)$ is $9 x^{2} + 9 y^{2} - 6 x = k$ . Then the value of $k =$

$a^{2} + b^{2}$

$a^{2} + b^{2} - 1$

$a^{2} + b^{2} + 1$

$0$

Q. Locus of the centroid of a triangle whose vertices are (1,0),(acost,asint),(bsint,−bcost) is 9x2+9y2−6x=k. Then the value of k=

a2+b2

a2+b2−1

a2+b2+1

0

Centroid is given by x=3x1​+x2​+x3​​ and y=3y1​+y2​+y3​​ So, x=31+acost+bsint​ and y=30+asint+(−bcost)​ ⇒(3x−1)=acost+bsint… (i) and 3y=asint−bcost… (2) Squaring and adding Eqs. (i) and (ii), we get 9x2−6x+1+9y2=a2+b2 ⇒9x2+9y2−6x=a2+b2−1 ∴k=a2+b2−1

Q. Locus of the centroid of a triangle whose vertices are $(1, 0), (a cos t, a sin t), (b sin t, - b cos t)$ is $9 x^{2} + 9 y^{2} - 6 x = k$ . Then the value of $k =$

$a^{2} + b^{2}$

$a^{2} + b^{2} - 1$

$a^{2} + b^{2} + 1$

$0$