If the minimum area of the triangle formed by a tangent to the ellipse $$\frac{x^2}{b^2} + \frac{y^2}{4a^2} = 1$$ and the co-ordinate axis is $$kab$$, then $$k$$ is equal to _________.

1. Parametric Form of the Tangent

The given ellipse is:

$$\frac{x^2}{b^2} + \frac{y^2}{(2a)^2} = 1$$

Any point on this ellipse can be represented in parametric form as:

$$(x_1, y_1) = (b \cos\theta, 2a \sin\theta)$$

The equation of the tangent at this point is given by:

$$\frac{x \cos\theta}{b} + \frac{y \sin\theta}{2a} = 1$$

2. Intercepts on Coordinate Axes

To find where the tangent cuts the axes:

X-intercept (put $$y = 0$$): $$x = \frac{b}{\cos\theta}$$

Y-intercept (put $$x = 0$$): $$y = \frac{2a}{\sin\theta}$$

3. Area of the Triangle

The area $$\Delta$$ of the right-angled triangle formed with the axes is:

$$\Delta = \frac{1}{2} \times \vert{}X\text{-intercept}\vert{} \times \vert{}Y\text{-intercept}\vert{}$$

$$\Delta = \frac{1}{2} \times \left\vert{}\frac{b}{\cos\theta}\right\vert{} \times \left\vert{}\frac{2a}{\sin\theta}\right\vert{}$$

$$\Delta = \frac{ab}{\vert{}\sin\theta \cos\theta\vert{}} = \frac{2ab}{\vert{}2\sin\theta \cos\theta\vert{}} = \frac{2ab}{\vert{}\sin 2\theta\vert{}}$$

4. Minimum Area Condition

For the area to be minimum, the denominator $$\vert{}\sin 2\theta\vert{}$$ must be maximum:

Maximum value of $$\vert{}\sin 2\theta\vert{} = 1$$

$$\Delta_{\min} = \frac{2ab}{1} = 2ab$$

Comparing this with the given minimum area $$kab$$:

$$kab = 2ab \implies k = 2$$

Answer:

The value of $$k$$ is $$2$$.