A curve C has the equation $$x^3 + 2xy - x - y^3 - 20 = 0$$ (a) Find \(\frac{dy}{dx}\) in terms of x and y: (b) Find an equation of the tangent to C at the point (3, -2), giving your answer in the form \(ax + by + c = 0\), where a, b and c are integers. - Edexcel - A-Level Maths Pure - Question 3 - 2014 - Paper 7

To find the equation of the tangent line at the point (3, -2), we first need to calculate (\frac{dy}{dx}) at this point.

Substituting (x = 3) and (y = -2) into the derivative we found:

$\frac{dy}{dx} = \frac{-3(3)^2 - 2(-2) + 1}{2(3) - 3(-2)^2}$

Calculating this:

• The numerator:
  • (-3(9) + 4 + 1 = -27 + 4 + 1 = -22)
• The denominator:
  • (6 - 3(4) = 6 - 12 = -6)

So we have:

$\frac{dy}{dx} = \frac{-22}{-6} = \frac{11}{3}$

Now we use the point-slope form of the line equation:

$y - y_1 = m(x - x_1)$

Substituting our slope and the point (3, -2):

$y + 2 = \frac{11}{3}(x - 3)$

Multiplying through by 3 to eliminate the fraction:

$3(y + 2) = 11(x - 3)$

Expanding this:

$3y + 6 = 11x - 33$

Rearranging gives:

$11x - 3y - 39 = 0$

This is in the required form (ax + by + c = 0), where a = 11, b = -3, and c = -39.