Question

find the equation tangent and normal to the curve 2x^3 +2y^3 - 9xy=0. at (2,1). ​

Answer 1

Answer:

5y-4x+3=0

Step-by-step explanation:

first of all diff. both sides wrt.x

6x²+6y²dy/dx-9(y+xdy/dx)=0

6x²+6y²dy/dx-9y-9xdy/dx=0

dy/dx(6y²-9xdy/dx) =9y-6x²

dy/dX=9y-6x²/6y²-9x

slope of tangent to the curve

dy/dx=9-24/6-18

=15/12=5/4

eqn of tangent

y-1 = 5/4(x-2)

5y-5-4x+8=0

5y-4x+3=0