Official solution
L1 : y2 = 4(x + 1)
Equation of tangent y = m(x + 1) + \[{1 \over m}\] ...(1)
L2 : y2 = 8(x + 2)
Equation of tangent y = m'(x + 2) + \[{2 \over {m'}}\]
\[ \Rightarrow \] y = m'x + 2\[\left( {m' + {1 \over {m'}}} \right)\] ....(2)
Since lines intersect at right angles
\[ \therefore \] mm' = -1
\[ \Rightarrow \] m' = \[{ - {1 \over {m}}}\]
Putting it in equation (2)
y = \[ - {1 \over m}x + 2\left( { - {1 \over m} - m} \right)\]
\[ \Rightarrow \] y = \[ - {1 \over m}x - 2\left( {m + {1 \over m}} \right)\] ....(3)
From equation (1) and (3)
m(x + 1) + \[{1 \over m}\] = \[ - {1 \over m}x - 2\left( {m + {1 \over m}} \right)\]
\[ \Rightarrow \] \[\left( {m + {1 \over m}} \right)x + 3\left( {m + {1 \over m}} \right)\] = 0
\[ \therefore \] x + 3 = 0