Question

If three real normals can be drawn to the parabola y² = 4ax from the point (a³, a) then-

• A. |a| <$\sqrt { 2 }$
• B. |a| < 1
• C. |a| >$\sqrt { 2 }$
• D. |a| > 1

Answer 1

Answer: (C)

|a| >$\sqrt { 2 }$

Answer 2

y = mx –2am – am³

am³ + 2am –a³m + a = 0

m₁ + m₂ + m₃ = 0 ..........(i)

m₁m₂ + m₂m₃ + m₃m₁ = .....(ii)

m₁m₂m₃ = – = – 1 .......(iii)

from (iii) m₁m₃ = –1/m₂

from (i) & (ii) – + m₁m₃ = (2 – a²)

– – = 2 – a²

– – 1 = (2 –a²)m₂

+ (2 – a²)m₂ + 1 = 0

so to have three solutions of the equation coeff. of m₂ should be –Ve so a² > 2 = |a| > $\sqrt { 2 }$

Alternatively

y = mx – 2am –am³ is satisfied by (a³, a) so

am³ + 2am – a³m + a = 0

m³ + (2–a²)m + 1 = 0

so 2 – a² < 0 so a² > 2 ; |a| >$\sqrt { 2 }$