Question

A curve y = f(x) passes through the point p(1, 1). The normal to the curve at p is a(y - 1) + (x - 1) = 0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, then the equation of the curve is

Answer 1

Step-by-step explanation:

Given A curve y = f(x) passes through the point p(1, 1). The normal to the curve at p is a(y - 1) + (x - 1) = 0. If the slope of the tangent at any point on the curve is proportional to the ordinate of the point, then the equation of the curve is

• So Normal to the curve is - 1
•                            y – 1 + x – 1 = 0
•                        So y = - x + 2
•                    So slope will be – 1
• Now normal to the curve is – dx / dy
•                   So – dx/dy (1,1) = - 1
• So dy / dx = 1
• Now dy/dx is proportional to ordinate
•                So dy/dx α y
•                     dy/dx = ky
•                    Or dy / y = k dx
•         Integrating we get
•                    So ∫ dy / y = ∫ k dx
•                     So log y = kx + c
•                So the point is (1,1)
•                 So log 1 = 0
•           So 0 = k + c
•             Or c = - k
•    So equation of curve is  
•                    log y = kx + c
•             Or log y = kx – k
•                            = k (x – 1)
•              Or y = e^k (x – 1)

Reference link will be

https://brainly.in/question/28891211

Answer 2

Answer:

upper one is Exactly correct