Question

find the equation of the tangent and normal to the following curve

y = 2x^2 +2y^2-9xy=0 at (2,1)​

Answer 1

Step-by-step explanation:

Given

y = 2x^2 +2y^2-9xy

p(1,2)

on differentiating above equation

dy/dx= 4x + 4ydy/dx - 9y -9xdy/dx

dy/dx(1 - 4y +9x)= 4x - 9y

tengent = dy/dx = (4x - 9y)/(1 - 4y + 9x)

putting the value of p

tangent = dy/dx = -1/15

for normal

Normal = 1/tengent

= - (1 - 4y + 9x)/(4x - 9y)