Question

Can anyone Solve this please

Answer 1

Answer:

Step-by-step explanation:

first find the derivative

y^2-2xy-4y-14x=0

2y dy/dx - 2y - 2x dy/dx - 4dy/dx - 14 = 0

dy/dx(2y-2x-4)=2y+14

dy/dx = y+7 / y-x-2

equation of normal

y-y1=-dx/dy(x-x1)

y-y1 = y-x-2/y+7 (x1-x)

y-3 = 3/10 * (-2-x)

10y-30 = -3x-6

-3x-10y+24=0

equation of tangent

y-y1/x-x1 =dy/dx

y-3/x+2 = 10/3

3y-9 = 10x+20

10x-3y+29=0