If \(\sqrt{(3) b x+a y=2 a b \text { touches the ellipse }\left(x^{2} / a^{2}\right)+\left(y^{2} / b^{2}\right)=1}\) then eccentric angle \(\theta\) of point of contact \(=\ldots \ldots \ldots\) (a) \((\pi / 2)\) (b) \((\pi / 3)\) (c) \((\pi / 4)\) (d) \((\pi / 6)\)

Given line equation: \(\sqrt{3}bx + ay = 2ab\) To rewrite it in the slope-intercept form (y = mx + c), first isolate y: \(ay = -\sqrt{3}bx + 2ab\) \(y = -\frac{\sqrt{3}b}{a}x + 2b\) Now we have the slope, m = -\(\frac{\sqrt{3}b}{a}\), and the y-intercept c = 2b.

Ellipse equation: \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) Parametric equations using eccentric angle θ: \(x = a \cos \theta\) \(y = b \sin \theta\)

Since the line touches the ellipse, the point of contact (x, y) must lie on both the line and the ellipse. Substitute the parametric equations into the line equation: \(b \sin \theta = -\frac{\sqrt{3}b}{a}(a \cos \theta) + 2b\) Solve for θ: \(\sin \theta = -\sqrt{3}\cos \theta + 2\) \(\sin \theta + \sqrt{3}\cos \theta = 2\) From the equation, we can notice that \(\sin \theta + \sqrt{3}\cos \theta = 2\) can be written as: \(2\sin(\theta + \frac{\pi}{3}) = 2\) So, \(\sin(\theta + \frac{\pi}{3}) = 1\) Thus, the eccentric angle θ of the point of contact is: \(\theta = \frac{\pi}{2} - \frac{\pi}{3}\) \(\theta = \frac{\pi}{6}\) Therefore, the correct answer for the eccentric angle θ is: (d) \(\frac{\pi}{6}\).