The locus of the centroid of the triangle who | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let the centroid of the triangle be $(x, y)$.Given vertices are $(x_1, y_1) = (a \cos t, a \sin t)$,$(x_2, y_2) = (b \sin t, -b \cos t)$,and $(x_3

Vedclass · Accepted Answer

$(3x - 1)^2 + (3y)^2 = a^2 + b^2$

Vedclass · Answer

(B) Let the centroid of the triangle be $(x, y)$.Given vertices are $(x_1, y_1) = (a \cos t, a \sin t)$,$(x_2, y_2) = (b \sin t, -b \cos t)$,and $(x_3, y_3) = (1, 0)$.The coordinates of the centroid are given by $x = \frac{x_1 + x_2 + x_3}{3}$ and $y = \frac{y_1 + y_2 + y_3}{3}$.Thus,$3x = a \cos t + b \sin t + 1$ and $3y = a \sin t - b \cos t$.Rearranging,we get $3x - 1 = a \cos t + b \sin t$ and $3y = a \sin t - b \cos t$.Squaring and adding both equations:$(3x - 1)^2 + (3y)^2 = (a \cos t + b \sin t)^2 + (a \sin t - b \cos t)^2$.Expanding the right side:$= a^2 \cos^2 t + b^2 \sin^2 t + 2ab \sin t \cos t + a^2 \sin^2 t + b^2 \cos^2 t - 2ab \sin t \cos t$.$= a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t) = a^2 + b^2$.Therefore,the locus is $(3x - 1)^2 + (3y)^2 = a^2 + b^2$.

The locus of the 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:

Similar Questions

If $L : ax + by + c = 0$ is a variable straight line,where $a, b$ and $c$ are the second,fourth,and seventh terms of an $AP$ respectively,then $L$ passes through the fixed point:

$A$ man starts walking from the point $P(-3, 4)$,touches the $x$-axis at $R$,and then turns to reach the point $Q(0, 2)$. The man is walking at a constant speed. If the man reaches the point $Q$ in the minimum time,then $50((PR)^{2} + (RQ)^{2})$ is equal to ..... .

$A$ variable line $\frac{x}{a}+\frac{y}{b}=1$ is such that $a+b=4$. The locus of the midpoint of the portion of the line intercepted between the axes is

$A$ straight line passing through a fixed point $(3, 5)$ intersects the coordinate axes at two points $A$ and $B$. If the locus of $C(x, y)$,which forms a rectangle with the points $A, O$ (origin),and $B$,is $ax + 2hxy + by = 0$,then $a + b + h =$

The locus of a point $P(x, y)$ which moves in such a way that the segment $OP$,where $O$ is the origin $(0, 0)$,has a slope of $\sqrt{3}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module