Question

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

• A. $(3x-1{)}^2+(3y{)}^2=a^2-b^2$
• B. $(3x-1{)}^2+(3y{)}^2=a^2+b^2$
• C. $(3x+1{)}^2+(3y{)}^2=a^2+b^2$
• D. $(3x+1{)}^2+(3y{)}^2=a^2+b^2$

Answer 1

The correct option is B $(3x-1{)}^2+(3y{)}^2=a^2+b^2$

We will call the centroid P(h,k). Later we will replace (h,k) with (x,y) after getting a relation between h and k. We will first find the centroid. We will equate it to (h,k) and eliminate the variable t.
The centroid of the given triangle is
$(\frac{acost+bsint+1}{3},\frac{asint-bcost+0}{3})$
This is equal to (h,k)
$\Rightarrow h=\frac{acost+bsint+1}{3}$
$\Rightarrow k=\frac{asint-bcost+0}{3}$
(3h-1)=a cost+b sint .....(1)
3k=a sint-b cost .....(2)
We have seen similar expressions before. We square and add them to eliminate t. $\Rightarrow$ $a^{2}co{s}^{2}t+b^{2}si{n}^{2}t+2abcostsint+a^{2}si{n}^{2}t+b^{2}co{s}^{2}t-2abcostsint.$
=$a^2+b^2$
We will replace (h,k) with (x,y)
$\Rightarrow (3x-1{)}^2+(3y{)}^2=a^2+b^2$