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Question

A curve y=f(x) passes through the point P(1,1). The normal to the curve at P is (y1)+(x1)=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
y=e2(x1)
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B
ey=e32x
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C
y=e2(1x)
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D
y=e(x1)
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Solution

The correct option is D y=e(x1)
Slope of the normal at (1,1)=1
So, the slope of the tangent at (1,1)=1
i.e., (dydx)(1,1)=1
Since dydx is proportional to y,
dydx=Ky
Here K=1(dydx)(1,1)=1
dyy=dx
logey=x+C
y=ex+C=Aex where A=ec
It passes through (1,1)
1=AeA=e1
y=e1ex=e(x1)

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