Find the values of $a$ and $b$ when the line tangent to the ellipse in the first quadrant forms a triangle whose area is a minimum given $a+b=1$. The problem has 3 parts to it. First I was asked to find the tangent line to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ in the first quadrant where the area of the triangle formed is a minimum. I found it to be $$\frac{x}{a\sqrt{2}}+\frac{y}{b\sqrt{2}}=1$$ The second part asks me to find the minimum area based from the first part of the question when $a+b=1$. This I found to be $\mbox{Area}=ab=b(1-b)$ since $a+b=1$ implies $a=1-b$. What I'm really puzzled about is the one that asks me to find the values of $a$ and $b$ for which this minimum area occurs. I'm not sure what to do with this. Please help.
P.S. The triangle formed here is the one bounded by the tangent line and the coordinate axes.
 A: Let's write the parametric equations of the ellipse as $P(t) = (a\cos t, b \sin t)$. The tangent line at the point $P(t)$ has equation:
$$
(b \cos t)x + (a \sin t)y = ab
$$ 
This line intersects the $y$-axis where $y = b/ \sin t$ and it intersects the $x$-axis where $x = a/ \cos t$. So, the area of the enclosed triangle is
$$
A(t) = \frac{ab}{2\sin t \cos t} = \frac{ab}{\sin (2t)}
$$
Clearly this has a minimum value of $ab$ when $t = \pi/4$. Setting $t = \pi/4$ in the tangent line equation above gives the same result that you found.
If $a + b = 1$, then the minimum area is $b(1-b)$. If you graph the function $f(b) = b(1-b)$, you will see that it's a "hump" shaped parabola that crosses the horizontal axis at $b=0$ and $b=1$. So, by setting $b$ to a value that's slightly larger than $0$ or slightly less than $1$, you can can make the area as small as you like. So, in some sense, the minimum area is zero (or, more precisely, it can be made arbitrarily small).
Note that the second part of the question is pretty much unrelated to the first part: it just deals with understanding the function $f(b) = b(1-b)$.
A: Using https://www.askiitians.com/iit-jee-coordinate-geometry/tangent-and-normal.aspx#equation-of-tangent-line-to-ellipse-in-different-forms,
the equation of any tangent at P$(a\cos t,b\sin t);0<t<\dfrac\pi2$ is
$$x\cos t/a+y\sin t/b=1$$
$$\iff \dfrac x{a\sec t}+\dfrac y{?}=1$$
We need to minimize $$\dfrac{ab}{\sin2t}\ge ab=a(1-a)$$
which can be made arbitrarily small by making $a\to0^+$ or $a\to1^-$
A: I  think area asked is between tangent and coordinate axis.
The tangent in 1st quadrant is $y= -mx + \sqrt{(a^2*m^2)+b^2}$  where m>0.
Y intercept of tangent is $\sqrt{(a^2*m^2) +b^2}$ ; 
X intercept is   $\sqrt{(a^2*m^2)+b^2}/m$
Area of triangle with coordinate Axes is $= ((a^2*m^2)+b^2)/2m$ ;
  Area = (1/2)(a^2*m+(b^2)/m)>=a*b>=1/4   I.e when m^2 = b^2/a^2
  and a is just greater than 1/2 and his just less than 1/2. In limiting case a and b are approaching towards 1/2.

