The equation of the tangent to the curve $y = \sqrt{9 - 3x^2}$ at the point where the ordinate and abscissa equal is...

Given curve is $y = \sqrt{9 - 3x^2}$

At the point where ordinate and abscissa are equal, $y = x$.

Substituting $y = x$ in the curve equation:

$x = \sqrt{9 - 3x^2}$

Squaring both sides:

$x^2 = 9 - 3x^2$

$4x^2 = 9$

$x = \dfrac{3}{2}$ (since $x \ge 0$ as $y \ge 0$)

Thus, the point of tangency is $\left(\dfrac{3}{2}, \dfrac{3}{2}\right)$.

Differentiating the curve equation $y^2 = 9 - 3x^2$ with respect to $x$:

$2y \dfrac{dy}{dx} = -6x$

$\dfrac{dy}{dx} = \dfrac{-3x}{y}$

At the point $\left(\dfrac{3}{2}, \dfrac{3}{2}\right)$, the slope of the tangent is:

$m = \dfrac{-3 \left(\dfrac{3}{2}\right)}{\dfrac{3}{2}} = -3$

The equation of the tangent is:

$y - \dfrac{3}{2} = -3\left(x - \dfrac{3}{2}\right)$

$2y - 3 = -6x + 9$

$6x + 2y - 12 = 0$

$3x + y - 6 = 0$

Answer: $3x + y - 6 = 0$