We are family

Area of ellipse $\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2} = 1$ is given by $A = \pi ab$ . Therefore, $A$ is minimum when $ab$ is minimum. For an ellipse we can express $\begin{cases} x = a\cos \theta \\ y = b \sin \theta \end{cases}$ . Since the ellipse passes through the point $(3.4)$ , $\begin{cases} a \cos \theta = 3 \\ b\sin \theta = 4 \end{cases}$ . $\implies ab\sin \theta \cos \theta = 12$ , $\implies ab = \dfrac {12}{\sin \theta \cos \theta} = \dfrac {24}{\sin 2\theta}$ .

Therefore, $\min (A) = \min (\pi ab) = \pi \min \left(\dfrac {24}{\sin 2\theta} \right) = \dfrac {24 \pi}{\max (\sin 2\theta)} = \boxed{24\pi}$ , when $\theta = \dfrac \pi 4$ .

Since the ellipse passes through $(3, 4)$ , the family of ellipses is $\frac{9}{a^2} + \frac{16}{b^2} = 1$ , which rearranges to $b = \frac{4a}{\sqrt{a^2 - 9}}$ for $a > b > 0$ .

Since the area of the ellipse is $A = \pi ab$ , it is $A = \frac{4\pi a^2}{\sqrt{a^2 - 9}}$ after substituting $b$ .

The minimum area $A$ occurs when the derivative $A' = \frac{4 \pi a(a^2 - 18)}{(a^2 - 9)^{\frac{3}{2}}} = 0$ and the second derivative $A'' > 0$ , which is at $a = 3\sqrt{2}$ for $a > 0$ .

If $a = 3\sqrt{2}$ , then $A = \frac{4\pi (3\sqrt{2})^2}{\sqrt{(3\sqrt{2})^2 - 9}} = \boxed{24\pi}$ .

We are required to minimise the function: $f =\pi ab$ subject to the constraint: $\frac{9}{a^2} + \frac{16}{b^2} = 1$

The Lagrangian can be constructed as follows:

$L = f + \lambda\left(\frac{9}{a^2} + \frac{16}{b^2}-1\right)$

Computing partial derivatives and equating to zero:

$\frac{\partial L}{\partial{a}} = \pi b - \frac{18\lambda}{a^3} = 0$ $\frac{\partial L}{\partial{b}} = \pi a - \frac{32\lambda}{b^3} = 0$ $\frac{\partial L}{\partial{\lambda}} = \frac{9}{a^2} + \frac{16}{b^2} - 1 = 0$

Multiplying the first equation by $a$ and the second by $b$ , adding them, and simplifying gives us: $\pi ab = \lambda$

Using this result in $\left(a\frac{\partial L}{\partial{a}} =0\right)$ and $\left(b\frac{\partial L}{\partial{b}} =0\right)$ and simplifying gives us:

$a^2 = 18$ $b^2 = 32$

From here, the required area can be computed to be $\boxed{\pi ab = 24\pi}$

To check whether the value obtained is a minimum, additional checks (second derivative tests) need to be carried out. This has not been demonstrated in this solution.

An alternate approach to the problem would be to avoid multi-variable calculus altogether. Consider the curve: $\frac{9}{a^2} + \frac{16}{b^2} = 1$ This can be plotted on the $a-b$ plane. Along with that, the rectangular hyperbola $ab = c$ can be plotted by varying c. This looks as such:

We see that when $ab = 24$ or $\pi ab = 24\pi$ , the hyperbola just touches the constraint graph. The miminum value can be computed by finding the hyperbola tangential to the constraint. This is in fact, in a way, a geometrical interpretation of the Lagrange multiplier approach. An interested reader is encouraged to look up relevant sources on the intuition behind Lagrange multipliers.