The tangent to the curve \(x y=25\) at any point on it cuts the coordinate axes at \(A\) and \(B\), then the area of the \(\triangle \mathrm{OAB}\) is

Given, curve \(x y=25 \quad \ldots (i)\)



On differentiating Eq. (i), w.r.t. ' \(x\) ', we get
\(x \frac{d y}{d x}+y=0\)
\(\Rightarrow \quad x \frac{d y}{d x}=-y\)
\(\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=-\frac{\mathrm{y}}{\mathrm{x}}\)
\(\Rightarrow \quad\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\operatorname{at}\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)}=\frac{-\mathrm{y}_{1}}{\mathrm{x}_{1}}\)
\(\therefore\) The equation of tangent at the point \(\left(x_{1}, y_{1}\right)\) is given by
\(y-y_{1}=\left(\frac{d y}{d x}\right)_{a t}\left(x_{1}, y_{1}\right)\) \(\Rightarrow \quad y-y_{1}=-\frac{y_{1}}{x_{1}}\left(x-x_{1}\right)\) \(\Rightarrow \quad x_{1} y-x_{1} y_{1}=-x_{1}+x_{1} y_{1}\) \(\Rightarrow \quad x_{1}+y_{1}=2 x_{1} y_{1}\) On dividing throughout by \(2 x_{1} y_{1}\), we get
\[
\frac{x}{2 x_{1}}+\frac{y}{2 y_{1}}=1
\]
Since, \(\left(x_{1}, y_{1}\right)\) satisfy the curve \(x y=25\)
[from Eq. (iii)]