If three real normals can be drawn to the parabola y 2 = 4ax from the point (a 3 , a) then-
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y = mx –2am – am 3
am 3 + 2am –a 3 m + a = 0
m 1 + m 2 + m 3 = 0 ..........(i)
m 1 m 2 + m 2 m 3 + m 3 m 1 = 
.....(ii)
m 1 m 2 m 3 = – 
= – 1 .......(iii)
from (iii) m 1 m 3 = –1/m 2
from (i) & (ii) – 
+ m 1 m 3 = (2 – a 2 )
– 
– 
= 2 – a 2
– 
– 1 = (2 –a 2 )m 2
+ (2 – a 2 )m 2 + 1 = 0
so to have three solutions of the equation coeff. of m 2 should be –Ve so a 2 > 2 = |a| > 
Alternatively
y = mx – 2am –am 3 is satisfied by (a 3 , a) so
am 3 + 2am – a 3 m + a = 0
m 3 + (2–a 2 )m + 1 = 0
so 2 – a 2 < 0 so a 2 > 2 ; |a| > 
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