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A curve y=f(x) passes through the origin...

A curve `y=f(x)` passes through the origin. Through any point `(x , y)` on the curve, lines are drawn parallel to the co-ordinate axes. If the curve divides the area formed by these lines and co-ordinates axes in the ratio `m : n ,` find the curve.

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The correct Answer is:
`y+sqrt(y^(2)-k^(2))=ce^(+-x//4)`

Area of OBPO: area of OPAP =`m:n`

`therefore (xy-int_(0)^(x)dx)/(int_(0)^(x)ydx)=m/n`
or `nxy=(m+n)int_(0)^(x)ydx`
Differentiating w.r.t. x, we get
`n(x(dy)/(dx)+y)=(m+n)y`
or `nx(dy)/(dx) = my` or `m/n(dx)/x=(dy)/y`
or `y=cx^(m//n)`
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