The equation of the tangent to the curve $y=\sqrt{9-2 x^2}$, at the point where the ordinate and abscissa…
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
$2 x+y+\sqrt{3}=0$
$2 x+y+3 \sqrt{3}=0$
$2 x-y-3 \sqrt{3}=0$
$2 x+y-3 \sqrt{3}=0$
Solution
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
$\begin{aligned}
& (y-\sqrt{3})=-2(x-\sqrt{3}) \\
& \text { i.e., } 2 x+y-3 \sqrt{3}=0
\end{aligned}$