1.

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 = eax–1(b)  y –1 = eax(c)  y = ea(x–1) (d)  None of these

Answer»

Correct option (c)  y = ea(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

⇒ loge y = ax+C

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

loge y = ax–a

⇒ y = ea(x–1)



Discussion

No Comment Found