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Question: Locus of centroid of the triangle whose vertices are (a cos t, sin t), (b sin t, -b cos t) and (1, 0...

Locus of centroid of the triangle whose vertices are (a cos t, sin t), (b sin t, -b cos t) and (1, 0) where t is a parameter is,
A. (3x+1)2+(3y)2=a2−b2{{\left( 3x+1 \right)}^{2}}+{{\left( 3y \right)}^{2}}={{a}^{2}}-{{b}^{2}}
B. (3x−1)2+(3y)2=a2−b2{{\left( 3x-1 \right)}^{2}}+{{\left( 3y \right)}^{2}}={{a}^{2}}-{{b}^{2}}
C. (3x−1)2+(3y)2=a2+b2{{\left( 3x-1 \right)}^{2}}+{{\left( 3y \right)}^{2}}={{a}^{2}}+{{b}^{2}}
D. (3x+1)2+(3y)2=a2+b2{{\left( 3x+1 \right)}^{2}}+{{\left( 3y \right)}^{2}}={{a}^{2}}+{{b}^{2}}

Explanation

Solution

Hint: We will be using the concept of coordinate geometry to solve the problem. We will first find the centroid of the triangle with the help of formula and then eliminate the parameter to find the locus of centroid of the triangle.

Complete step-by-step answer:
Now, we have the three vertices of the triangle as, (a cos t, a sin t) (b sin t, -b cos t) and (1, 0).
Now, we let the coordinate of the centroid of the triangle as (x, y). Now, we know that the coordinate of centroid of a triangle with coordinate of vertices as,(x1,y1),(x2,y2) and (x3,y3) is (x1+x2+x33,y1+y2+y33)\left( {{x}_{1}},{{y}_{1}} \right),\left( {{x}_{2}},{{y}_{2}} \right)\ and\ \left( {{x}_{3}},{{y}_{3}} \right)\ is\ \left( \dfrac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}}{3},\dfrac{{{y}_{1}}+{{y}_{2}}+{{y}_{3}}}{3} \right)
So, we have from the given coordinates of vertices the coordinate of centroids,
x=acos⁡t+bsin⁡t+13.........(1) y=asin⁡t−bcos⁡t+03.........(2) \begin{aligned} & x=\dfrac{a\cos t+b\sin t+1}{3}.........\left( 1 \right) \\\ & y=\dfrac{a\sin t-b\cos t+0}{3}.........\left( 2 \right) \\\ \end{aligned}
Now, we have to eliminate sin t, cos t from (1) and (2) and form a relation between x and y to find the locus of centroid.
Now, we have from (1) and (2),
3x−1=acos⁡t+sin⁡t 3y=asin⁡t−bcos⁡t \begin{aligned} & 3x-1=a\cos t+\sin t \\\ & 3y=a\sin t-b\cos t \\\ \end{aligned}
Now, we square and add both of the above equations, so we have,
(3x−1)2+(3y)2=(acos⁡t+bsin⁡t)2+(asin⁡t−bcos⁡t)2{{\left( 3x-1 \right)}^{2}}+{{\left( 3y \right)}^{2}}={{\left( a\cos t+b\sin t \right)}^{2}}+{{\left( a\sin t-b\cos t \right)}^{2}}
Now, we will use algebraic identity that,
(a+b)2=a2+b2+2ab (a−b)2=a2+b2−2ab \begin{aligned} & {{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab \\\ & {{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab \\\ \end{aligned}
So, we have,
(3x−1)2+(3y)2=a2cos⁡2t+b2sin⁡2t+2abcos⁡tsin⁡t+a2sin⁡2t+b2cos⁡2t−2abcos⁡tsin⁡t =a2(sin⁡2t+cos⁡2t)+b2(sin⁡2t+cos⁡2t) \begin{aligned} & {{\left( 3x-1 \right)}^{2}}+{{\left( 3y \right)}^{2}}={{a}^{2}}{{\cos }^{2}}t+{{b}^{2}}{{\sin }^{2}}t+2ab\cos t\sin t+{{a}^{2}}{{\sin }^{2}}t+{{b}^{2}}{{\cos }^{2}}t-2ab\cos t\sin t \\\ & ={{a}^{2}}\left( {{\sin }^{2}}t+{{\cos }^{2}}t \right)+{{b}^{2}}\left( {{\sin }^{2}}t+{{\cos }^{2}}t \right) \\\ \end{aligned}
Now, we know that,
sin⁡2t+cos⁡2t=1{{\sin }^{2}}t+{{\cos }^{2}}t=1
So, we have,
(3x−1)2+(3y)2=a2+b2{{\left( 3x-1 \right)}^{2}}+{{\left( 3y \right)}^{2}}={{a}^{2}}+{{b}^{2}}
Hence, the correct option is (C).

Note: To solve these types of questions it is important to note the way we have eliminated parameter t from our equation it is also important to note that we have first formed a relation between given parameter and the coordinate of point whose locus we have to find and then eliminated the parameter t.