find the equation of the tangent line to the curve at the given point. \n\nx²/9 + y²/4 = 2 at (3, 2)\n\na. y…

Explanation:

Step1: Differentiate the given equation implicitly.

Differentiate $\frac{x^{2}}{9}+\frac{y^{2}}{4}=2$ with respect to $x$. Using the power - rule and chain - rule, we have $\frac{2x}{9}+\frac{2y}{4}y' = 0$.

Step2: Solve for $y'$.

First, simplify the equation to $\frac{2x}{9}+\frac{y}{2}y' = 0$. Then, isolate $y'$: $\frac{y}{2}y'=-\frac{2x}{9}$, so $y'=-\frac{4x}{9y}$.

Step3: Find the slope of the tangent line at the point $(3,2)$.

Substitute $x = 3$ and $y = 2$ into $y'$: $y'|_{(3,2)}=-\frac{4\times3}{9\times2}=-\frac{2}{3}$.

Step4: Use the point - slope form of a line to find the equation of the tangent line.

The point - slope form is $y - y_{1}=m(x - x_{1})$, where $(x_{1},y_{1})=(3,2)$ and $m = -\frac{2}{3}$. $y - 2=-\frac{2}{3}(x - 3)$. Expand the right - hand side: $y - 2=-\frac{2}{3}x+2$. Add 2 to both sides to get $y=-\frac{2}{3}x + 4$.

Answer:

C. $y =-\frac{2}{3}x + 4$