If $\mathrm{P}\left(a, b\right)$ is the point to which the origin is to be shifted by translation of axes so as to remove the first degree terms from the equation $4x^2+2xy+y^2-8x-4y-12=0$ and $\theta$ is the angle through which the axes are to be rotated about the origin so as to remove the $xy$-term from the above equation, then $a+b+3\tan 2\theta =$

Given that origin is shifted at $P\left(a,b\right)$ so as to remove the first degree term from $4x^2+2xy+y^2-8x-4y-12=0$.

Let $\left(X,Y\right)$ be new coordinates

$\therefore x=X+a \& y=Y+b$

Now put $x=X+a$ and $y=Y+b$in equation$4x^2+2xy+y^2-8x-4y-12=0$, we get

$4{\left(X+a\right)}^2+2\left(X+a\right)\left(Y+b\right)+{\left(Y+b\right)}^2-8\left(Y+a\right)-4\left(Y+b\right)-12=0$

$\Rightarrow 4X^2+4a^2+8Xa+2XY+2bX+2aY+2ab+Y^2+b^2+3Yb-8X-8a-4Y-4b-12=0$

$\Rightarrow 4X^2+Y^2+2XY+\left(8a+2b-8\right)X+\left(2b+2a-4\right)Y+4a^2+b^2+2ab-8a-4b-12=0 ...\left(1\right)$

Removing first degree term, we get

$8a+2b-8=0$ and  $2b+2a-4=0$

i.e.  $4a+b=4 ...\left(2\right)$  and  $a+b=2 ...\left(3\right)$

Now equation $\left(1\right)$ becomes,

$4X^2+Y^2+2XY+4a^2+b^2+2ab-8a-4b-12=0 ....\left(4\right)$

Now the axes are to be rotated through an angle $\theta$ about the origin to remove $xy$ term from the equation.

Let $X=x^{\text{'}}\cos \theta -y^{\text{'}}\sin \theta \text{\&} Y=x^{\text{'}}\sin \theta +y^{\text{'}}\cos \theta$

Substituting $X=x^{\text{'}}\cos \theta -y^{\text{'}}\sin \theta \& Y=x^{\text{'}}\sin \theta +y^{\text{'}}\cos \theta$ in $\left(4\right)$, we get

$4{\left(x^{\text{'}}\cos \theta -y^{\text{'}}\sin \theta \right)}^2+{\left(x^{\text{'}}\sin \theta +y^{\text{'}}\cos \theta \right)}^2+2\left(x^{\text{'}}\cos \theta -y^{\text{'}}\sin \theta \right)\left(x^{\text{'}}\sin \theta +y^{\text{'}}\cos \theta \right)+4a^2+b^2+2ab-8a-4b-12=0$

$\Rightarrow {x^{\text{'}}}^2\left(4{\cos }^2\theta +{\sin }^2\theta +2\sin \theta \cos \theta \right)+{y^{\text{'}}}^2\left(4{\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta \right)+x^{\text{'}}{y}^{\text{'}}\left(-8\sin \theta \cos \theta +2\sin \theta \cos \theta +2\left({\cos }^2\theta -{\sin }^2\theta \right)\right)+4a^2+b^2+2ab-8a-4b-12=0$

Removing $x^{\text{'}}{y}^{\text{'}}$ term, we get

$-6\sin \theta \cos \theta +2\left({\cos }^2\theta -{\sin }^2\theta \right)=0$

$\Rightarrow -3\sin 2\theta +2\cos 2\theta =0$

$\Rightarrow \tan 2\theta =\frac{2}{3}$

Now,

$a+b+3\tan 2\theta =2+3·\frac{2}{3}=4$