A curve `y=f(x)` passes through 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

Options

A. `y = e^(ax)-1`
B. `y=e^(ax)+1`
C. `y=e^(ax)+a`
D. `y= e^(a(x-1))`

Answer

d

Explanation

` :' ` Equation of normal at P (1,1) is
` ay + x = a +1 ` [ given ]
` :' " Slope of norml at " (1,1) = -1/a " " ` …(i)
Also , given `(dy)/(dx) prop y rArr (dy)/(dx) = ky `
` [ (dy)/(dx)]_(1,1) = k =a` [ from Eq. (i) ]
Then,` (dy)/(dx) = ay `
` rArr (dy)/y = a dx`
` rArr In | y | = ax + C`
` :' " It is passing through (1,1) ,then C = -a`
` rArr In |y| = a (x-1)`
` rArr |y| = e^(a(x-1))`