How to find the locus of this?

Question

Here, in Q4, how to find locus? I took one vertex as $(a,0)$ and other as $(0,b)$ and centroid as $(x, y)$. I am getting these two equations and am unable to eliminate $a$ and $b$ completely-

Answer 1

If one endpoint of the hypotenuse is $A=(t,0)$, then the other endpoint is $B=\big(0,\sqrt{2-t^2}\big)$ and the midpoint of $AB$ is then $M={1\over2}\big(t,\sqrt{2-t^2}\big)$.

If $G$ is the triangle centroid, then $MG$ is perpendicular to $AB$ and $MG=\sqrt2/6$. It follows that the coordinates of $G$ are: $$ G=\left( {1\over2}t+{1\over6}\sqrt{2-t^2},{1\over2}\sqrt{2-t^2}+{1\over6}t \right), $$ which is then the parametric equation of the locus.

To eliminate $t$, just notice that the coordinates of $G$ satisfy: $$ 3y-x={4\over3}\sqrt{2-t^2} \quad\hbox{and}\quad 3x-y={4\over3}t. $$ Squaring both equalities and adding the results gives the stated equation.