Step 1:
Tangent and normal at $(2,-1)$ to the hyperbolic $9x^2-5y^2=31$
The tangent at $(x_1,y_1)$ is $9xx_1-5yy_1=31$
Here $(x_1,y_1)=(2,-1)$.
Therefore the tangent is $9\times 2x-5\times (-1)y=31$
$\Rightarrow 18x+5y=31$
Step 2:
The normal is $\perp$ to the tangent and its equation is of the form $5x-18y+k=0$
It passes through $(2,-1)$.
Therefore $10+18+k=0$
$\Rightarrow k=-28$
The equation of the normal is $5x-18y-28=0$
Step 3:
Equation of tangent : $18x+5y-31=0$
Equation of normal : $5x-18y-28=0$