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(D) Given the curve $y = \sqrt{9 - 3x^2}$. At the point where abscissa $(x)$ and ordinate $(y)$ are equal, $x = y$.Substitute $y = x$ into the equatio

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$3x + y - 6 = 0$

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(D) Given the curve $y = \sqrt{9 - 3x^2}$. At the point where abscissa $(x)$ and ordinate $(y)$ are equal, $x = y$.Substitute $y = x$ into the equation: $x = \sqrt{9 - 3x^2} \implies x^2 = 9 - 3x^2 \implies 4x^2 = 9 \implies x^2 = \frac{9}{4} \implies x = \frac{3}{2}$ (since $y$ must be positive).So, the point of tangency is $(\frac{3}{2}, \frac{3}{2})$.Differentiating $y^2 = 9 - 3x^2$ with respect to $x$: $2y \frac{dy}{dx} = -6x \implies \frac{dy}{dx} = -\frac{3x}{y}$.At $(\frac{3}{2}, \frac{3}{2})$, the slope $m = -\frac{3(3/2)}{3/2} = -3$.The equation of the tangent is $y - y_1 = m(x - x_1) \implies y - \frac{3}{2} = -3(x - \frac{3}{2})$.$y - \frac{3}{2} = -3x + \frac{9}{2} \implies 3x + y - 6 = 0$.

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

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