The transformed equation of $3x^2+4xy+y^2-8x-4y-4=0$ is $f\left(X,Y\right)=a{\mathrm{X}}^2+2h\mathrm{XY}+b{\mathrm{Y}}^2+c=0$

Given $3x^2+4xy+y^2-8x-4y-4=0$

Now let $x=X+h y=Y+k$ and put in given equation we get,

$\Rightarrow 3{\left(X+h\right)}^2+4\left(X+h\right)\left(Y+k\right)+{\left(Y+k\right)}^2-8\left(X+h\right)-4\left(Y+k\right)-4=0$

$\Rightarrow 3X^2+Y^2+4XY+X\left(6h+4k-8\right)+Y\left(4h+2k-4\right)+4hk+k^2-8h-4k-4=0$

$$Now making coefficient of $X \& Y$ to be zero we get,

 $6h+4k-8=0 ...\left(1\right)$

$4h+2k-4=0 ...\left(2\right)$

Now solving equation $\left(1\right) \& \left(2\right)$ we get $h=0 \& k=2$

Now putting the value of $h \& k$ in new shifted equation we get,

$f\left(X,Y\right)=3X^2+Y^2+4XY-8$

Then $f\left(1,1\right)=0$