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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 (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
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(x−1) 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=Ae⇒A=e−1 ∴y=e−1ex=e(x−1)

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