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. 5 x squared plus 3 xy plus 3 y squared equals 44​; ​(2​,2​)

Answer 1

Answer:

y = 13*( -x/9 + 1/5)

Step-by-step explanation:

Given:

- The curve has an equation as follows:

                               $44 = 5x^2 + 3xy + 3y^2$

Find:

a. Verify that the given point (2​,2​) lies on the curve.

b. Determine an equation of the line tangent to the curve at the given point.

Solution:

- To verify whether the point lies on the given curve we will substitute the coordinates of the point into the equation as follows:

                              44 = 5*(2)^2 + 3*(2)(2) + 3*(2)^2

                              44 = 20 + 12 + 12

                              44 = 44 ......Hence proven.

- The equation of the line tangent to the curve is expressed as a linear function as follows:

                              y = m*x + C

Where, m is the gradient of the line.

            C is the y-intercept.

                              m = Δy / Δx = dy/dx

- We will take the derivative of the given curve with respect to x as follows:

                             $0 = 10x + 3*( y + xy' ) + 6y*y'\\\\-10x - 3y = y' ( 3x + 6y)\\\\ y' = - \frac{10x + 3y}{3x + 6y}$

- Evaluate y' at the point (2,2) we get:

                            y' = - ( 10(2) + 3(2) ) / ( 3(2) + 6(2) )

                            y' = - ( 26 ) / (18)

                            y'= m = - 13/9

- To evaluate C, we will use the point (2,2) for linear expression above with m as follows:

                            y = -13*x/9 + C

                            2 =-13*(2)/9 + C

                            C = 13 / 5

- The equation of the tangent is as follows:

                            y = 13*( -x/9 + 1/5)