The equation of a tangent to the hyperbola $16x^2-25y^2-96x+100y-356=0$, which makes an angle $\frac{\pi }{4}$with the transverse axis, is

We have hyperbola $16x^2-25y^2-96x+100y-356=0$

This can be rewritten as

$\begin{array}{l}\frac{(x-3{)}^2}{25}-\frac{(y-2{)}^2}{16}=1\\ \end{array}$

Equation of tangent to (i) is 

$y-2=m(x-3)+\sqrt{a^2{m}^2-b^2}$

Clearly,

$\begin{array}{l}m=\tan {45}^{\circ }=1,a^2=25\text{ and }{b}^2=16\\ \\ \Rightarrow y-2=x-3+3\end{array}$

i.e. y = x + 2 is the required equation of tangent