Find $a$, $b$ such that the ellipse $(x/a)^2 + (y/b)^2 = 1$ passes through $(\sqrt 2, 2)$ and has minimum area

Question

I am working on a problem in which, for $a$, $b \gt 0$, we let $(x/a)^2 + (y/b)^2 = 1$ describe an ellipse.
I am required to use the method of Lagrange multipliers and the corresponding second derivative test to find $a$ and $b$ such that the ellipse passes through $(x, y) = (\sqrt 2, 2)$ and such that the area of the ellipse $A = \pi ab$ is minimised amongst all such ellipses.

My thoughts:

Firstly, since $(\sqrt 2, 2)$ lies on the ellipse, we must have that

$$\frac{2}{a^2} + \frac{4}{b^2} = 1.$$

Secondly, I let $g(a, b) = \pi ab$.
With the aim of minimising this function, I calculated $g_a = \pi b$ and $g_b = \pi a$ and set them both equal to $0$.
However, this only gives that $a = b = 0$, which is not allowed.

I am not yet sure how to progress further with this question and would appreciate hints.

Answer 1

Your constraint function is $$ f(a,b) = \frac{2}{a^2} + \frac{4}{b^2} - 1 $$ and your minimization function is $g(a,b) = \pi ab$

Using the Lagrange multiplier method, this requires $\nabla g = \lambda \nabla f $, thus we have the following system

$$ \pi b = -\lambda\frac{4}{a^3} $$ $$ \pi a = -\lambda\frac{8}{b^3} $$ $$ f(a,b) = 0 $$

Answer 2

HINT: use the Lagrange Multiplier method $$F=\pi ab+\lambda(\frac{2}{a^2}+\frac{4}{b^2}-1)$$