The point to which the origin should be shifted in order to eliminate the x and y terms from the equation 9…

Let the origin be shifted at $\left(h,k\right)$

Let $\left(x^{\text{'}},y^{\text{'}}\right)$be the new co-ordinate

$\therefore x=x^{\text{'}}+h$& $y=y^{\text{'}}+k$

$\because$Given equation is

$9x^2+4y^2+10x+12y+1=0$

Put $x=x^{\text{'}}+h$&   $y=y^{\text{'}}+k$

$\Rightarrow 9{\left(x^{\text{'}}+h\right)}^2+4{\left(y^{\text{'}}+k\right)}^2+10\left(x^{\text{'}}+h\right)+12\left(y^{\text{'}}+k\right)+1=0$

$\Rightarrow 9{\left(x^{\text{'}}\right)}^2+9h^2+18x^{\text{'}}h+4{\left(y^{\text{'}}\right)}^2+4k^2+8y^{\text{'}}k+10x^{\text{'}}+10h+12y^{\text{'}}+12k+1=0$

$\Rightarrow 9{\left(x^{\text{'}}\right)}^2+4{\left(y^{\text{'}}\right)}^2+x^{\text{'}}\left(18h+10\right)+y^{\text{'}}\left(8k+12\right)+9h^2+4k^2+10h+12k+1=0$

So, to eliminate $x^{\text{'}} \& y^{\text{'}},$put

$18h+10=0 \& 8k+12=0$

$\therefore h=\frac{-5}{9} \& k=-\frac{3}{2}$

$\therefore$Origin should be shifted at $\left(\frac{-5}{9},\frac{-3}{2}\right)$