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.
For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$,  the equation of the tangent line at $(a\cos t, b \sin t)$ is given by
$$\frac{\cos t}ax+\frac{\sin t}by=1$$
which intersects the axes at $\frac a{\cos t}$ and $\frac b{\sin t}$, respectively. Then, the area of the triangle is
$$Area = \frac12 \frac a{\cos t}\frac b{\sin t}=\frac{ab }{\sin2t}\ge ab 
$$
Here, $a=\frac1{\sqrt2}$ and $b=1$, allowing the smallest area $\frac1{\sqrt2}$.
Let $(h,k)$ be the point on ellipse from where you draw the tangent, then,
$$2h^2+k^2=1$$
You already know the slope of tangent is $$\frac{dy}{dx}=\frac{-2h}{k}$$
Constructing the equation of hypotenuse from this,
$$\frac{y-k}{x-h}=\frac{-2h}{k}$$
Putting $y=0$ , you get $x=a$,
$$a=\frac{1}{2h}$$
Similarly $$b=\frac{1}{k}$$
You need to minimise the area, i.e. , $$=\frac{ab}{2}$$
$$=\frac{1}{4hk}$$
$$=\frac{1}{4\cdot{k}\cdot{}\sqrt{\frac{1-k^2}{2}}}$$
$$=\frac{1}{2\sqrt{2}\cdot{k}\cdot{}\sqrt{1-k^2}}$$
Can you continue?