JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

1 Easy

1 of 1

Application +4 | -1

Locus of the centroid of triangle whose vertices are

\left( {a\cos \;t,\;a\sin t} \right),\left( {b\sin t, - b\cos t} \right)

and (1,0), where t is a parameter, is

View answer and explanation

Correct option: B

{\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}

Option B (

{\left( {3x - 1} \right)^2} + {\left( {3y} \right)^2} = {a^2} + {b^2}

) is correct because it represents the locus of the centroid derived from the given vertices.

More Questions from 3D GEOMETRY SYSTEM

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\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

\frac{x-1}{1}=\frac{2-y}{2}, z=3