How would I do this question?

a) 3

b) S: (½, -15)

Step-by-step explanation:

a) gradient of the normal at R:

4y + x = 24

y = (-¼)x + 24

Slope of the Normal: -¼

Slope of the tangent at R:

-1 ÷ -¼ = 4

dy/dx = 2kx - 8

4 = 2k(2) - 8

12 = 4k

k = 3

Slope of the tangent: 4

Coordinates of R: (2,y)

y = 3(2)² - 8(2) - 5 = -9

Equation of tangent:

y = 4x + c

-9 = 4(2) + c

c = -17

y = 4x - 17

Point of intersection of tangent and the other curve:

4x - 17 = 4x - (1/x³) - 9

4x⁴ - 17x³ = 4x⁴ - 1 - 9x³

-8x³ = -1

x³ = ⅛

x = ½

y = 4(½) - 17

y = -15

S: (½, -15)

10000000000000000000000000000000