Correct option is (d) y=x+2

Explaination::

tangents to the curve y^{2}=16x is y=mx+4/m , so it must satisfy xy=-4

\(x({mx+\frac{4}{m}})=-4\)

\(mx^2+\frac{4}{m}x+4=0 ,\)

since it has equal roots , therefore D=0

\(\frac{16}{m^2}-16m=0\)

\(m^3=1 \)

m=1

therefore , the equation of commom tangent is y=x+4 i.e. x-y+4=0