Please help me with this

Question

Find the equation of the tangent line to the curve at the given point using implicit differentiation.

Trisectrix of Maclaurin

x(x² + y²) = 3x² − y²

 at (1, −1)

Answer 1

x(x² + y²) = 3x² − y² [apply the derivative operator to get:

The derivative of x(x² + y²) is found using the product rule (f'g + fg') which becomes 1(x² + y²) + x(2x + 2y•dy/dx).

The derivative of the right side is 6x - 2y•dy/dx. so the entire implicit differentiation is:

1(x² + y²) + x(2x + 2y•dy/dx).= 6x - 2y•dy/dx [now solve for dy/dx as follows:

x² + y² + 2x² + 2xy•dy/dx.= 6x - 2y•dy/dx

3x² + y² + 2xy•dy/dx.= 6x - 2y•dy/dx

2xy•dy/dx + 2y•dy/dx.= 6x - 3x² - y²

dy/dx(2xy + 2y) = (6x - 3x² - y²)

dy/dx = (6x - 3x² - y²)/(2xy + 2y)

At x = 1 and y = -1 we get:

dy/dx = (6(1) - 3(1)² - (-1)²)/(2(1)(-1) + 2(-1))

dy/dx = 2/-4 = -1/2 [this is the slope "m" of the tangent line

So, using the point-slope form of a line we get:

y - (-1) = (-1/2)(x - 1)

y + 1 = -1/2x + 1/2

y = -1/2x - 1/2