A tangent to the ellipse meets the $x$ and $y$ axes . If $O$ is the origin, find the minimum area of triangle $AOB$. 

The question is that : A tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the $x$ and $y$ axes respectively at $A$ and $B$. If $O$ is the origin, find the minimum area of triangle $AOB$.


What I have attempted:
Because the equation of an ellipse is $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$$
We have that the $x$ intercept is $x=\frac{a^2}{x_1}$ and $y$ intercept is $y=\frac{b^2}{y_1}$
Hence the area of a triangle $(AOB)$ is given by $A=\frac{a^2b^2}{2x_1y_1}$ now I am stuck, how should I proceed?
 A: $A = \frac {a^2 b^2}{2xy}$
constrained by
$\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$
$\frac{dA}{dx} = \frac {-2a^2b^2 (y + x y')}{(2xy)^2} = 0\\
y + x y' = 0\\
y' = -\frac yx$
Differentiating the constraint.
$\frac x{a^2} + \frac {y y'}{b^2} = 0\\
y' = -\frac {b^2x}{a^2y}\\
\frac {b^2x}{a^2y} = \frac yx\\
b^2 x^2 = a^2 y^2$
Again from the constraint:
$b^2 x^2 + a^2 y^2 = a^2 b^2\\
2a^2 y^2 = a^2 b^2\\
y^2 = \frac {b^2}{2}\\
y = \frac b{\sqrt {2}}\\
2xy = ab\\
A = \frac {a^2b^2}{2xy} = ab$
alternate
$x = a \cos t\\
y = b \sin t\\
A = \frac {ab}{2} \csc t \sec t\\
\frac{dA}{dt} =  \frac {ab}{2} \sec t \csc t ( -\cot t + \tan t) = 0\\
\cot t = \tan t\\
t = \frac {\pi}{4}\\
A = ab$
That is much nicer.
A: With the change of coordinates $x=au, y=bv$ the problem boils down to the same problem where the original ellipse is replaced by a unit circle. In such a case it is trivial that the minimum area of $AOB$ is $1$. Since the determinant of the Jacobian of the linear map $x=au, y=bv$ is $ab$, the answer to the original question is $\color{red}{ab}$, plain and simple.
