Let a tangent be drawn to the ellipse $$\frac{x^2}{27} + y^2 = 1$$ at $$(3\sqrt{3}\cos\theta, \sin\theta)$$ where $$\theta \in \left(0, \frac{\pi}{2}\right)$$. Then the value of $$\theta$$ such that the sum of intercepts on axes made by this tangent is minimum is equal to :

The ellipse is $$\frac{x^2}{27} + y^2 = 1$$ and the point of tangency is $$(3\sqrt{3}\cos\theta, \sin\theta)$$. The equation of the tangent at this point is $$\frac{x \cdot 3\sqrt{3}\cos\theta}{27} + \frac{y \cdot \sin\theta}{1} = 1$$, which simplifies to $$\frac{x\cos\theta}{3\sqrt{3}} + y\sin\theta = 1$$.

The x-intercept is found by setting $$y = 0$$: $$x = \frac{3\sqrt{3}}{\cos\theta}$$. The y-intercept is found by setting $$x = 0$$: $$y = \frac{1}{\sin\theta}$$. The sum of intercepts is $$S(\theta) = \frac{3\sqrt{3}}{\cos\theta} + \frac{1}{\sin\theta} = 3\sqrt{3}\sec\theta + \csc\theta$$.

To minimize, we differentiate and set equal to zero: $$S'(\theta) = 3\sqrt{3}\sec\theta\tan\theta - \csc\theta\cot\theta = 0$$. This gives $$3\sqrt{3}\frac{\sin\theta}{\cos^2\theta} = \frac{\cos\theta}{\sin^2\theta}$$, so $$3\sqrt{3}\sin^3\theta = \cos^3\theta$$, hence $$\tan^3\theta = \frac{1}{3\sqrt{3}} = \frac{1}{\sqrt{27}}$$, giving $$\tan\theta = \frac{1}{\sqrt{3}}$$, and therefore $$\theta = \frac{\pi}{6}$$.