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2015-09-15T18:47:55+05:30

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Curve  y = log x    we take the base  as  e  (not decimal)
slope of tangent =  dy/dx  = 1 / x
           at (1,  0)  slope =  1

so slope of normal at  (1,0) =  -1  as the product of slopes is -1..

equation of normal is  y = - x + c
   as it passes throug h  (1,0),  c = 1
         so normal is :  y = 1 - x
     y-intercept  of this  normal is :  1

Area of the traingle formed by the normal to the curve, and the axes is
       1/2 * 1 * 1 = 1/2
 
1 5 1
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