The area of a triangle formed by a tangent to the curve $2 x y=a^2$ and the coordinate axes, is

$a^2$

Let $P(x_1, y_1)$ be a point on the curve

$2 x y=a^2$                  .....(i)

Then,

$2 x_1 y_1=a^2$                  .....(ii)

Now,

$2 x y=a^2 \Rightarrow 2\left(x \frac{d y}{d x}+y\right)=0 \Rightarrow \frac{d y}{d x}=-\frac{y}{x} \Rightarrow\left(\frac{d y}{d x}\right)_P=-\frac{y_1}{x_1}$

The equation of the tangent to (i) at $P(x_1, y_1)$ is

$y-y_1=-\frac{y_1}{x_1}\left(x-x_1\right)$

$\Rightarrow x_1 y-x_1 y_1=-x y_1+x_1 y_1$

$\Rightarrow x y_1+y x_1=2 x_1 y_1 \Rightarrow x y_1+y x_1=a^2$      [Using (ii)]

This tangent meets the coordinate axes at

$A\left(a^2 / y_1, 0\right)$ and $B\left(0, a^2 / x_1\right)$

∴  Area of $\triangle O A B=\frac{1}{2} \times O A \times O B$

⇒  Area of $\triangle OAB=\frac{1}{2} \times \frac{a^2}{y_1} \times \frac{a^2}{x_1}=\frac{a^4}{2 x_1 y_1}=a^2$          [Using (i)]