4. The curve C has equation 4x^2 – y3 – 4xy + 2y = 0. P has coordinates (–2, 4) lies on C. (a) Find the exact value of d d y x at the point P. (6) The normal to C at P meets the y-axis at the point A. (b) Find the y coordinate of A

Differentiate Implicitly with respect to x- ( do it term by term, terms with only x's diff normally, terms containing a Y should be differentiated following the standard pattern but then multiplied by dy/dx as an unknown)d/dx(4x^2)+ (d(y^3)/dy * dy/dx) +d/dx(-4xy)+d/dx(2^y)=08x - 3y^2(dy/dx) -4y -4x(dy/dx) + ln(2).2^y(dy/dx)= 0 (collect dy/dx terms and factorise)dy/dx = (8x -4y)/(3y^2 +4x -ln2.2^y) (plug in the coordinates of P for x and y)dy/dx = (-32)/(40-16ln(2))This is grad of tangent and the answer to (a)For (b) the grad of normal is (-1)/(dy/dx)grad normal = (-40 +16ln(2))/(-32) use ( y-y₁ = M(x - x₁) )4-A=(-40 + (16ln(2))/-32) ( -2 -0 )A = 13/2 -ln(2)