Question

If $A(a,0,0)$ and $B(0,b,0)$ are two vertices of a triangle whose third vertex lies on the curve $y^2=2x-3z$, then the locus of the centroid of the triangle is

• A. $9y^2-6x-6by+9az+b^2+2a=0$
• B. $9y^2-6x-6by+9z+b^2+2a=0$
• C. $9y^2-6bx-6y+9z+b^2+2a=0$
• D. $9y^2-6bx-6y+9az+b^2+2a=0$

Answer 1

The correct option is B $9y^2-6x-6by+9z+b^2+2a=0$

Let the coordinates of centriod is $(h,k,l)$
Given: $A(a,0,0),B(0,b,0),C(x,y,z)$
We know,
$h=\frac{a+0+x}{3}\Rightarrow x=3h-a$
$k=\frac{0+b+y}{3}\Rightarrow y=3k-b$
$l=\frac{0+0+z}{3}\Rightarrow z=3l$
Given $(x,y,z)$ lies on the curve $y^2=2x-3z$.
$\Rightarrow (3k-b{)}^2=2(3h-a)-3(3l)$
$\Rightarrow 9k^2+b^2-6kb=6h-2a-9l$
$\Rightarrow 9k^2-6h-6bk+9l+b^2+2a=0$
Locus of centroid is:
$\therefore 9y^2-6x-6by+9z+b^2+2a=0$