Question

a. verify that the given point lies on the curve. b. determine an equation of the line tangent to the curve at the given point. 3x^{2}+7xy + 6y^{2}=93, (3,2) a. the point (3,2) lies on the curve. the equation of the tangent line is y = (use integers or fractions for any numbers in the equation.) b. the point (3,2) does not lie on 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 \(y =-\frac{32}{45}x+\frac{62}{15}\)