a. verify that the given point lies on the curve. b. determine an equation of the line tangent to the curve…

Explanation:

Step1: Check if point lies on curve

Substitute (x = 3) and (y=2) into (3x^{2}+7xy + 6y^{2}). [3\times3^{2}+7\times3\times2+6\times2^{2}=3\times9 + 42+6\times4=27 + 42+24=93] Since the result is 93, the point ((3,2)) lies on the curve.

Step2: Differentiate the equation implicitly

Differentiate (3x^{2}+7xy + 6y^{2}=93) with respect to (x). Using the power - rule and product - rule: [6x+7y + 7x\frac{dy}{dx}+12y\frac{dy}{dx}=0]

Step3: Solve for (\frac{dy}{dx})

[7x\frac{dy}{dx}+12y\frac{dy}{dx}=-6x - 7y] [\frac{dy}{dx}(7x + 12y)=-6x - 7y] [\frac{dy}{dx}=\frac{-6x - 7y}{7x+12y}]

Step4: Find the slope of the tangent at ((3,2))

Substitute (x = 3) and (y = 2) into (\frac{dy}{dx}): [\frac{dy}{dx}=\frac{-6\times3-7\times2}{7\times3 + 12\times2}=\frac{-18-14}{21 + 24}=\frac{-32}{45}]

Step5: Find the equation of the tangent line

Use the point - slope form (y - y_{1}=m(x - x_{1})), where ((x_{1},y_{1})=(3,2)) and (m =-\frac{32}{45}). [y - 2=-\frac{32}{45}(x - 3)] [y-2=-\frac{32}{45}x+\frac{32}{15}] [y=-\frac{32}{45}x+\frac{32}{15}+2] [y=-\frac{32}{45}x+\frac{32 + 30}{15}] [y=-\frac{32}{45}x+\frac{62}{15}]

Answer:

A. The point ((3,2)) lies on the curve. The equation of the tangent line is (y =-\frac{32}{45}x+\frac{62}{15})