Step 1:

Equation of the rectangular hyperbola is $xy=12$.

Equation of the tangent to $xy=12$ at $(x_1,y_1)$ is $\large\frac{1}{2}$$(xy_1+yx_1)$$=12$

Here $(x_1,y_1)=(3,4)$

Therefore the tangent is $4x+3y=24$

Step 2:

The normal is $\perp$ to the tangent and the equation of the form $3x-4y=k$.It passes through $(3,4)$

Therefore $9-16=k$

$\Rightarrow k=-7$

The normal is $3x-4y=-7$

$\Rightarrow 3x-4y+7=0$

Step 3:

Equation of tangent : $4x+3y=24$

Equation of normal : $3x-4y+7=0$