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A curve y=f(x) passes through point P...

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 (a) `( b ) (c) y=( d ) e^(( e ) (f) K(( g ) (h) x-1( i ))( j ))( k ) (l)` (m) (b) `( n ) (o) y=( p ) e^(( q ) (r) K e (s))( t ) (u)` (v) (c) `( d ) (e) y=( f ) e^(( g ) (h) K(( i ) (j) x-2( k ))( l ))( m ) (n)` (o) (d) None of these

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The correct Answer is:
`y = e^(a(x-1)), (1+(e^(-a))/(a) - (1)/(2a))`
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