<s>[INST] 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.
 [/INST] You have made a small error. You wrote that
$$m = \dfrac{b-0}{0-a}$$ but then said that this is equal to
$$ \dfrac{-a}{b}$$
This is clearly wrong. In fact,
$$m=-\frac{b}{a}$$

The Solution
We have $y=-\frac{b}{a}x+b=\frac{ab-bx}{a}$. Substituting this into $2x^2+y^2=1$ yields
$$\begin{align}
&2x^2+\left(\frac{ab-bx}{a}\right)^2=1\\
\implies& 2a^2x^2+b^2x^2-2ab^2x+a^2b^2=a^2\\
\implies& x^2(2a^2+b^2)+x(-2ab^2)+(a^2b^2-a^2)=0\\
\end{align}$$
We could then go on to try to solve this quadratic, but in fact we can do better. Note that the tangent line will intersect the ellipse once, and only once. Hence, the discriminant of this quadratic in $x$ is equal to $0$:
$$4a^2b^4-4(2a^2+b^2)(a^2b^2-a^2)=0$$
We can divide through by $a^2$, as we don't have a degenerate triangle with $a=0$, simplifying to
$$\begin{align}&4b^4-4(2a^2+b^2)(b^2-1)=0\\
\implies&4b^4-4(2a^2b^2-2a^2+b^4-b^2)=0\\
\implies & a^2(8b^2-8)=4b^2\\
\implies & a^2=\frac{b^2}{2(b^2-1)}
\end{align}$$
Hence, we have, assuming without loss of generality that $a,b\geqslant 0$, that
$$a=\frac{b}{\sqrt{2(b^2-1)}}$$
So we find that the area of the triangle, which we know is equal to $\frac{1}{2}ab$, is in fact equal to
$$\frac{b^2}{2\sqrt{2(b^2-1)}}$$
So we now have an expression for the area, $A$, in terms of one variable only! This means that to find the minimum value of $A$, we can just differentiate with respect to $b$, solve for stationary points, check it is a minimum, and we are basically done!

I hope that was helpful. Thank you for giving me the opportunity to attack this interesting question; I'd never come across a similar one before. What I find particularly pleasing about this solution is that we have no need to look at the derivative of the ellipse, which would've made things rather complicated. If you have any questions please don't hesitate to ask!
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