Question

A curve f(x) has normal at the point P(1,1) given by a(y – 1) +(x – 1) = 0. If the slope of tangent at any point on the curve is proportional to the ordinate at that point, then the curve is

(a)  y = e^ax–1

(b)  y –1 = e^ax

(c)  y = e^a(x–1) 

(d)  None of these

Answer 1

Correct option (c)  y = e^{a(x – 1)} 

Explanation:

Normal at point P is ay + x = a + 1 

Slope of tangent at P = a = (dy/dx) (1, 1) ...(1)

Now dy/dx α ⇒ dy/dx = ky  ⇒ (dy/dx)_(1,1) = k ...(2)

From (1) & (2) k = a

dy/dx  = ay ⇒ dy/y  = a.dx

⇒ log_e y = ax+C

Now C = – a (as curve passes through (1,1))

log_e y = ax–a

⇒ y = e^a(x–1)