The equation of the tangent to the curve \(y=\sqrt{9-2 x^2}\), at the point where the ordinate and abscissa are equal, is

Given curve is \(y=\sqrt{9-2 x^2}\)
If ordinate and abscissa are equal, we get \(y=x\).
\(\therefore \quad\) Equation of the curve becomes \(x^2=9-2 x^2\) \(\Rightarrow x= \pm \sqrt{3}\)
If \(x=-\sqrt{3}\), then \(y=\sqrt{9-2(3)}=\sqrt{3}\)
In this case, \(x \neq y\).
Hence, \(x \neq-\sqrt{3}\)
\(\therefore \quad x=\sqrt{3}\) and \(y=\sqrt{3}\)
\(\therefore \quad\) Slope of the tangent to the given curve is
\(\begin{aligned}
& 2 y \frac{\mathrm{d} y}{\mathrm{~d} x}=-4 x \\
\therefore \quad & \text { at }(\sqrt{3}, \sqrt{3}), \frac{\mathrm{d} y}{\mathrm{~d} x}=-2
\end{aligned}\)
\(\therefore \quad\) Equation of the required tangent is
\(\begin{aligned}
& (y-\sqrt{3})=-2(x-\sqrt{3}) \\
& \text { i.e., } 2 x+y-3 \sqrt{3}=0
\end{aligned}\)