Given equations:\[\] 1. \( y^2 = 4a(x + c) \) \[\] 2. \( y^2 = 4b x \) \[\] Equation of Normal For the first equation, the normal equation is: \[ y = m(x + c) - \left(am^3 + 2am\right). \] For the second equation, the normal equation is: \[ y = mx - \left(bm^3 + 2bm\right). \] Common Normal Condition If the two equations share a common normal, then: \[ mx - \left(bm^3 + 2bm\right) = m(x + c) - \left(am^3 + 2am\right). \] Simplifying: \[ mx - bm^3 - 2bm = mx + mc - am^3 - 2am. \] Cancel out \( mx \) from both sides: \[ -bm^3 - 2bm = mc - am^3 - 2am. \] Rearranging: \[ (a - b)m^3 + m(2a - 2b) = mc. \] Factoring out \( m \): \[ m\left((a - b)m^2 + 2a - 2b - c\right) = 0. \] Solutions for \( m \) 1. \( m = 0 \): This corresponds to a vertical normal, which is trivial. 2. Solve for \( m^2 \): \[ (a - b)m^2 + 2a - 2b - c = 0. \] Rearranging: \[ m^2 = \frac{-(2a - 2b - c)}{a - b}. \] Taking the square root: \[ m = \pm \sqrt{\frac{c - 2(a - b)}{a - b}}. \] Condition for Common Normal For \( m \) to be real, the discriminant must be non-negative: \[ c - 2(a - b) > 0. \] Thus: \[ c > 2(a - b). \]