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