Question

If $\theta$ is parameter, $A=(a\sec \theta ,b\tan \theta )andB=(-a\tan \theta ,b\sec \theta ),O=(0,0)$ then the locus of the centroid of $△OAB$ is

• A. $9xy=ab$
• B. $xy=9ab$
• C. $x^2-9y^2=a^2-b^2$
• D. $x^2-y^2=\frac{1}{9}(a^2-b^2)$

Answer 1

The correct option is A $9xy=ab$

Let the centroid be $G(x,y)$,

then

$G=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

$G(x,y)=(\frac{asect-atant}{3},\frac{btant+bsect}{3})$

$\therefore x=\frac{asect-atant}{3},y=\frac{bsect+btant}{3}$

$⟹3x=asect-atnat-\left(1\right)$

$⟹3y=bsect+btant‘-\left(2\right)$

Multiply eqs.(1) with (2)-

$3x\times 3y=(asect-atant)(bsect+btant)⟹9xy=ab(sect-tant)(sect+tant)⟹9xy=ab(se{c}^{2}t-ta{n}^{2}t)$

$9xy=ab\left(1\right)⟹9xy=ab$