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
that point, determine the equation of the curve. Also, obtain the area
bounded by the y-axis, the curve and the normal to the curve at P
​

Answer 1

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.

the slope of the tangent at any point on the curve is proportional to the ordinate of that point,

To Find : Equation of the curve.  

the area bounded by the y-axis, the curve and the normal to the curve at P

​

Solution:

slope of the tangent at any point on the curve is proportional to the ordinate of that point,

=> dy/dx ∝  y

=> dy/y  =  kdx

k is a constant

integrating both sides

=> ln y  = kx  + c

passes through (1 , 1)

=> 0 = k + c

=> c = - k

=> ln y  = kx   - k

=> ln y = k(x - 1)

or $y = e^{k(x-1)}$

normal to the curve at P is a (y - 1) + (x - 1) = 0

=> x + y = 2

=> slope = - 1

Slope of tangent   = 1

dy/dx  = ky  =  1

y = 1

=> k(1) = 1

=> k = 1

Hence   y = e⁽ˣ⁻¹⁾

equation of the curve is  y = e⁽ˣ⁻¹⁾

y = e⁽ˣ⁻¹⁾

Area between curve and x axis  

= ∫ydx = ∫e⁽ˣ⁻¹⁾dx  x from 0 to 1

=> e⁽ˣ⁻¹⁾   x from 0 to 1

Area between curve and x axis    =   e⁽¹⁻¹⁾ - e⁽⁰⁻¹⁾ = 1 - 1/e

= 0.63212 sq units

Area by trapezium formed = (1/2) (2 + 1) * 1 = 3/2   ( ref figure)

or by  ∫(-x + 2)dx  from 0 to 1   as  y = - x + 2 equation of line

= (-x²/2 + 2x)   x from 0 to 1  = -1/2 + 2 = 3/2 = 1.5

 area bounded by the y-axis, the curve and the normal to the curve at P

= 1.5 - 0.63212

= 0.86788 sq units

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