Step-by-step Solution:

Let the chord touch the parabola at two points \( P \) and \( Q \). \[ t_2 = -t_1 - \frac{2}{t_1} \] Since it subtends an angle of \(90^\circ\) at the origin, the product of slopes at the origin is \(-1\). \[ m_1 \times m_2 = -1 \] Using the slope formula for a parabola \( y^2 = 4ax \): \[ m_1 = \frac{2a t_1}{a t_1^2}, \quad m_2 = \frac{2a t_2}{a t_2^2} \] Since their product is \(-1\): \[ t_1 \times t_2 = -4 \] From the equation: \[ -4 = t_1 \left( -t_1 - \frac{2}{t_1} \right) \] \[ -4 = -t_1^2 - 2 \] \[ t_1^2 = -2 + 4 = 2 \] \[ t_1 = -\sqrt{2} \] Thus, \[ t_2 = 2\sqrt{2} \] The points on the parabola corresponding to these parameters are: \[ P(at_1^2, 2a t_1), \quad Q(at_2^2, 2a t_2) \] \[ P(2a, -2\sqrt{2}a), \quad Q(8a, 4\sqrt{2}a) \] The slope of the chord line is: \[ m = \frac{6\sqrt{2}a}{6a} \] \[ m = \sqrt{2} \]