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Question

# Tangent to a curve y=f(x) intersects the yâˆ’axis at a point P. A line perpendicular to this tangent through P passes through another point (1, 0). The differential equation of the curve is

A
ydydxx(dydx)2=1
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B
xd2ydx2+(dydx)2=0
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C
ydydx+x=1
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D
xdydx+(dydx)2=1
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Solution

## The correct option is A ydydx−x(dydx)2=1The equation of the tangent at the point R(x, f(x)) is Y−f(x)=f′(x)(X−x) The coordinates of the point P are (0, f(x)−xf′(x)) The slope of the perpendicular line through P is f(x)−xf′(x)−00−1=−1f′(x) ⇒f(x)f′(x)−x(f′(x))2=1 ⇒ydydx−x(dydx)2=1 which is the required differential equation of the curve y=f(x)

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