= 1 will be a normal to the parabola y2 = 4ax if, for some value of m, it is identical with y = mx - 2am - am3 i.e. mx - y = (2am + am3) Comparing coefficients, we get
m
1∕p
=
−1
1∕q
=
2a,m+am3
1
⇒ mp = - q ∴ m = −
q
p
and mp = m (2a + am2) or P = 2a + am2 = 2a + a (−