Locus of the centroid of a triangle whose vertices are \((1,0) .(a \cos t, a \sin t)\), \((b \sin t,-b \cos…
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 \(k\) is equal to
\(a^2+b^2\)
\(a^2+b^2-1\)
\(a^2+b^2+1\)
0
Solution
Centroid is given by
\(x=\frac{x_1+x_2+x_3}{3} \text { and } y=\frac{y_1+y_2+y_3}{3}\)
So, \(x=\frac{1+a \cos t+b \sin t}{3}\)
and \(y=\frac{0+a \sin t+(-b \cos t)}{3}\)
\(\Rightarrow(3 x-1)=a \cos t+b \sin t\) ...(i)
and \(3 y=a \sin t-b \cos t\) ...(ii)
Squaring and adding Eqs. (i) and (ii), we get
\(\begin{aligned}
& 9 x^2-6 x+1+9 y^2=a^2+b^2 \\
& \Rightarrow 9 x^2+9 y^2-6 x=a^2+b^2-1 \\
& \therefore \quad k=a^2+b^2-1 \\
\end{aligned}\)