The pole of the line 2x-y-4=0 with respect to the parabola  x²-4x-8y+12=0 is the point

 a.(10,2) 

b.(3,4)

c.(12,10)

d.(3,2)

Let the pole of the line be (x₀,y₀).Then its eqn. is

S₀=0 i.e; xx₀-2(x+x₀)-4(y+y₀)+12=0

(x₀-2)x-4y-(2x₀+4y₀-12)=0

Comparing this with given eqn.,we get,

x₀-2=8 and 2x₀+4y₀-12=16

So,x₀=10 and y₀=2.Hence,the point is (10,2).