Smallest possible area for triangle I'm solving the following question:

The two legs of a right triangle lie along the positive x and y axes.
The hypotenuse is tangent to the ellipse $2x^2 + y^2 = 1$. What is the
smallest possible area for such a triangle ?

This is my attempt at solving the problem:
Let $a$ be the length of the x axis and $b$ be the length in it's y axis if the triangle. So it's area is $1/2 a.b$
Let's find the tangent of the ellipse:
$2x^2 + y^2 = 1$
$4x + 2y \dfrac{dy}{dx} = 0$
$\dfrac{dy}{dx} = \dfrac{-2x}{y}$
Since the two points on the triangle are $(0,b)$ and $(a,0)$, we can
find the slope of the line:
$m = \dfrac{b-0}{0-a} = \dfrac{-a}{b}$
So the equation of hypotonuse is $y = \dfrac{-a}{b}x + c$
Since $(0,b)$ is one of the point in the line, we can substitute it in
the above line to find that $b = c$. Solving it with the other
co-ordinate, we can find the following fact $b = a$.
So now area is $1/2 a.b = 1/2 a^2$
Now I used the first derivative test to find that the function attains
it's local minima at $0$. So the smallest possible area for such a
triangle is zero.
I have missed lots of steps in my above solution for brevity,
but I can expand if needed. My question: is zero
the right answer ?  Unfortunately, the book I'm using doesn't provide
the answer to this question.


Outline : We will write down the equation of tangent at arbitrary point $P$ of ellipse. Let it intersect the coordinate axes in $A$ and $B$. We identify the legs of the right triangle formed can be identified as $x$- and $y$- intercepts of the tangent line. So we find the smallest area of triangle minimizing the product of the intercepts.
Let $P=(h,k)$. Then tangent at $P$ is $2xh+ky=1$ which can be written as
$$\frac{x}{1/(2h)}+\frac{y}{1/k}=1$$
Area $=\frac{1}{2}\cdot \frac{1}{(2h)}\cdot \frac{1}{k}\cdot$ where $2h^2+k^2=1$
By AM-GM,
$$\frac{2h^2+k^2}{2} \ge \sqrt{2h^2\cdot k^2}$$
$$\Rightarrow \frac{1}{hk} \ge \frac{2\sqrt{2}}{2h^2+k^2}=2\sqrt{2}$$
$$\therefore A_{\text{min}}=\frac{1}{4hk}=\frac{2\sqrt{2}}{4}=\frac{1}{\sqrt{2}}$$
Equality occurs at $2h^2=k^2$ which can be used to determine coordinates of optimal $P$.
