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 _________.

We have the ellipse whose equation is $$\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{4a^{2}}=1\;.$$

Let a line having slope $$m$$ touch this ellipse. A convenient way to write a line of given slope is

$$y=mx+c\,,$$

where $$c$$ is its intercept on the y-axis. In order that this line be a tangent to the ellipse, after substituting it in the ellipse the resulting quadratic in $$x$$ must have equal roots. In other words, its discriminant must be zero.

Substituting $$y=mx+c$$ in the ellipse:

$$\dfrac{x^{2}}{b^{2}}+\dfrac{(mx+c)^{2}}{4a^{2}}=1.$$

Multiplying every term by $$4a^{2}b^{2}$$ to clear the denominators, we get

$$4a^{2}x^{2}+b^{2}(mx+c)^{2}=4a^{2}b^{2}.$$

Expanding the square and arranging in descending powers of $$x$$,

$$\bigl(4a^{2}+b^{2}m^{2}\bigr)x^{2}+2b^{2}mc\,x+\bigl(b^{2}c^{2}-4a^{2}b^{2}\bigr)=0.$$

This quadratic in $$x$$ will have equal roots when its discriminant $$D$$ is zero. The discriminant formula is

$$D=(\text{coefficient of }x)^{2}-4(\text{coefficient of }x^{2})(\text{constant term}).$$

Applying it,

$$[\,2b^{2}mc\,]^{2}-4\bigl(4a^{2}+b^{2}m^{2}\bigr)\bigl(b^{2}c^{2}-4a^{2}b^{2}\bigr)=0.$$

Dividing by the common factor $$4$$ gives

$$b^{4}m^{2}c^{2}-\bigl(4a^{2}+b^{2}m^{2}\bigr)\bigl(b^{2}c^{2}-4a^{2}b^{2}\bigr)=0.$$

Expanding the right-hand product and simplifying, every term containing $$c^{2}$$ can be collected on one side to obtain

$$c^{2}\bigl[\,b^{2}(4a^{2}+b^{2}m^{2})-b^{4}m^{2}\bigr]=4a^{2}b^{2}(4a^{2}+b^{2}m^{2}).$$

Inside the square brackets, the two $$b^{4}m^{2}$$ terms cancel, leaving

$$c^{2}(4a^{2}b^{2})=4a^{2}b^{2}(4a^{2}+b^{2}m^{2}).$$

Because $$a\neq0,\,b\neq0,$$ we can divide by $$4a^{2}b^{2}$$ and get

$$c^{2}=4a^{2}+b^{2}m^{2}\;.$$

Thus the equation of the tangent with slope $$m$$ is

$$y=mx+\sqrt{\,4a^{2}+b^{2}m^{2}\,}\;.$$

We are interested only in the tangent that meets both co-ordinate axes in the first quadrant, i.e. the one whose slope is negative. Therefore we take $$m<0$$ and write

$$c=\sqrt{\,4a^{2}+b^{2}m^{2}\,}\;.$$

The intercept on the x-axis is obtained by setting $$y=0$$:

$$0=mx+c\quad\Longrightarrow\quad x=-\dfrac{c}{m}\;.$$

Because $$m<0$$ and $$c>0$$, this $$x$$-intercept is indeed positive. The intercept on the y-axis is simply

$$y=c\;.$$

Hence the triangle formed with the axes has base $$-\dfrac{c}{m}$$ and height $$c$$. The area $$\Delta$$ of a right-angled triangle is $$\tfrac12(\text{base})(\text{height})$$, so

$$\Delta=\dfrac12\left(-\dfrac{c}{m}\right)c=-\dfrac{c^{2}}{2m}\;.$$

Replacing $$c^{2}$$ by its expression found earlier,

$$\Delta(m)=-\,\dfrac{4a^{2}+b^{2}m^{2}}{2m}\;.$$

To locate the minimum value of $$\Delta$$ we differentiate with respect to $$m$$ (remembering $$m<0$$):

Write the function in a simpler split form first:

$$\Delta(m)=-\dfrac{4a^{2}}{2m}-\dfrac{b^{2}m}{2}= -\dfrac{2a^{2}}{m}-\dfrac{b^{2}m}{2}\;.$$

Differentiate term by term with respect to $$m$$:

$$\dfrac{d\Delta}{dm}= -2a^{2}\left(-\dfrac1{m^{2}}\right)-\dfrac{b^{2}}{2}= \dfrac{2a^{2}}{m^{2}}-\dfrac{b^{2}}{2}\;.$$

For an extremum, this derivative must be zero:

$$\dfrac{2a^{2}}{m^{2}}-\dfrac{b^{2}}{2}=0\quad\Longrightarrow\quad \dfrac{2a^{2}}{m^{2}}=\dfrac{b^{2}}{2}\;.$$

Cross-multiplying,

$$4a^{2}=b^{2}m^{2}\quad\Longrightarrow\quad m^{2}=\dfrac{4a^{2}}{b^{2}}\;.$$

Since we insisted $$m<0,$$ we take

$$m=-\dfrac{2a}{b}\;.$$

We now compute the corresponding intercept $$c$$:

$$c^{2}=4a^{2}+b^{2}m^{2}=4a^{2}+b^{2}\left(\dfrac{4a^{2}}{b^{2}}\right)=4a^{2}+4a^{2}=8a^{2}\;.$$

Hence $$c=2\sqrt2\,a\;.$$

The minimum area is therefore

$$\Delta_{\min}=-\,\dfrac{c^{2}}{2m}=-\,\dfrac{8a^{2}}{2\left(-\dfrac{2a}{b}\right)}=\dfrac{8a^{2}}{\dfrac{4a}{b}}=2ab\;.$$

The question states that the minimum area equals $$kab$$, and we have just obtained $$2ab$$. Therefore

$$k=2\;.$$

Hence, the correct answer is Option 2.