Log_2aa=x    then, a=(2a)^x  ......(1)

Log_3a2a=y    then,2a=(3a)^y ......(2)

Log_4a 3a=z  then, 3a=(4a)^z ......(3)

So, 

a=(2a)^x  [from (1)]

Or, a=(3a)^xy    [from(2)]

Or, a=(4a)^xyz     [from(3)]

Multiplying both sides by 4a,

4a.a=4a.(4a)^xyz  

Or,(2a)² =(4a)^{xyz + 1} 

Or,(3a)^2y =(4a)^xyz+1 

Or,(4a)^2yz =(4a)^xyz+1 

Or, 2yz = xyz+1 .proved.

Here, the given equation of parabola is y²= 8x.

The equation of tangent to the parabola y²=8x is,

y= mx + 2/m

This tangent passes through the point (-2, 3)

So, 3 = -2m + 2/m

or, 3m + 2m² = 2

or, 2m²+3m - 2= 0

or, 2m² + (4 - 1)m -2 = 0

or, 2m² + 4m - m - 2 = 0

or, 2m(m + 2) - 1(m+2) = 0

or, (m + 2) (2m - 1) = 0

Either, Or,

m = -2 m = 1/2

Required angle is,