Question

If $A(4,-3),B(0,2)$ and points $B$ lies on the locus $Y=1+x^2$, find the equation of the locus of the centroid of $△ABC$.

Answer 1

Let the coordinate of c be (a, b) and the coordinates of the centroid of the triangle ABC be (h, k) then,

$(h,k)=(\frac{5+2+a}{3},\frac{-2+4+b}{3})$

$(h,k)=(\frac{a+7}{3},\frac{b+2}{3})$

$h=\frac{a+7}{3}$

$h^2={\left(\frac{a+7}{3}\right)}^2$

$9h^2=a^2+14a+49$ ……$\left(1\right)$

$k=\frac{b+2}{3}$

$k^2={\left(\frac{b+2}{3}\right)}^2$

$9k^2=b^2+4b+4$ ………$\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$

$9h^2+9k^2=a^2+14a+49+b^2+4b+4$

$9h^2+9k^2=a^2+b^2+14a+4b+53$

Replacing h and k by x and y we get

$9x^2+9y^2=a^2+b^2+14a+4b+53$ is the required equation.