Question

Find locus of centroid of a triangle whose vertices are given by (t,t²), (1,0), (t², 2t +1) where t is real parameter.

• A. 9x² - 18xy + 9y² - 6x + 3y + 1 = 0
• B. 9x² + 18xy + 9y² - 6x + 3y + 1 = 0
• C. 9x² - 18xy + 9y² + 6x - 3y + 1 = 0
• D. 9x² + 18xy + 9y² + 6x - 3y + 1 = 0

Answer 1

Answer: (A)

9x² - 18xy + 9y² - 6x + 3y + 1 = 0

Answer 2

• The vertices of the triangle are given as $A = (t, t^2)$, $B = (1, 0)$, and $C = (t^2, 2t + 1)$.
• Let the centroid of the triangle be $G = (x, y)$. The coordinates of the centroid are given by the average of the coordinates of the vertices: $x = \frac{x_1 + x_2 + x_3}{3}$ $y = \frac{y_1 + y_2 + y_3}{3}$
• Substituting the coordinates of the vertices: $x = \frac{t + 1 + t^2}{3}$ $y = \frac{t^2 + 0 + (2t + 1)}{3}$
• This gives us the parametric equations for the locus of the centroid:
  1. $3x = t^2 + t + 1$
  2. $3y = t^2 + 2t + 1$
• Subtract equation (1) from equation (2) to eliminate $t^2$: $3y - 3x = (t^2 + 2t + 1) - (t^2 + t + 1)$ $3(y - x) = t$
• Substitute $t = 3(y - x)$ back into equation (1): $3x = (3(y - x))^2 + (3(y - x)) + 1$ $3x = 9(y - x)^2 + 3(y - x) + 1$ $3x = 9(y^2 - 2xy + x^2) + 3y - 3x + 1$ $3x = 9y^2 - 18xy + 9x^2 + 3y - 3x + 1$
• Rearrange the terms to form the equation of the locus: $9x^2 - 18xy + 9y^2 - 6x + 3y + 1 = 0$
• The discriminant $B^2 - 4AC = (-18)^2 - 4(9)(9) = 0$, indicating the locus is a parabola.