use implicit differentiation to find an equation of the tangent line to the cur\n$x^{2}+6xy + 12y^{2}=28$…

Explanation:

Step1: Differentiate both sides

Differentiate (x^{2}+6xy + 12y^{2}=28) with respect to (x). Using the sum rule ((u + v+w)^\prime=u^\prime + v^\prime+w^\prime), where (u = x^{2}), (v=6xy), (w = 12y^{2}). For (u=x^{2}), (u^\prime=2x). For (v = 6xy), use the product rule ((uv)^\prime=u^\prime v+uv^\prime) (here (u = 6x), (v = y)), so (v^\prime=6y+6x\frac{dy}{dx}). For (w = 12y^{2}), use the chain rule ((f(g(x)))^\prime=f^\prime(g(x))\cdot g^\prime(x)), so (w^\prime=24y\frac{dy}{dx}). The derivative of the right - hand side is (0). So (2x+6y + 6x\frac{dy}{dx}+24y\frac{dy}{dx}=0).

Step2: Solve for (\frac{dy}{dx})

Group the terms with (\frac{dy}{dx}): (6x\frac{dy}{dx}+24y\frac{dy}{dx}=-2x - 6y). Factor out (\frac{dy}{dx}): (\frac{dy}{dx}(6x + 24y)=-2x - 6y). Then (\frac{dy}{dx}=\frac{-2x - 6y}{6x + 24y}=\frac{-x - 3y}{3x + 12y}).

Step3: Evaluate (\frac{dy}{dx}) at the point ((2,1))

Substitute (x = 2) and (y = 1) into (\frac{dy}{dx}): (\frac{dy}{dx}\mid_{(x = 2,y = 1)}=\frac{-2-3\times1}{3\times2+12\times1}=\frac{-5}{18}).

Step4: Use the point - slope form (y - y_{1}=m(x - x_{1}))

Here (x_{1}=2), (y_{1}=1), (m =-\frac{5}{18}). (y - 1=-\frac{5}{18}(x - 2)). Multiply through by (18): (18y-18=-5x + 10). Rearrange to get (5x+18y=28).

Answer:

(5x + 18y=28)