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Question

# The curve y = f(x) is such that the area of the trapezium formed by the coordinate axes, ordinate of an arbitrary point and the tangent at this point equals half the square of its abscissa. The equation of the curve can be

A
y=cx2±x
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B
y=cx2±1
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C
y=cx±x2
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D
y=cx2±x2±1
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Solution

## The correct option is A y=cx2±xLet P(x, y) be any point on the curve. Length of intercept on y-axis by any tangent at P is OT=y−xdydx ∴ Area of trapezium OLPTO=12(PL+OT)OL =12(y+y−xdydx)x=12(2y−xdydx)x According to question Area of trapezium OLPTO=12x2 i.e.,12(2y−xdydx)x=±12x2⇒2y−xdydx=±x or dydx−2yx=±1 Which is linear differential equation and I.F.=e−2In x=1x2 ∴ The solution is yx2=∫1x2dx+c=±1x+c ∴y=±x+cx2 or y=cx2 or y=cx2±x, where c is an arbitrary constant

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