The equation of the curve which is such that the portion of the axis of `x` cut off between the origin and tangent at any point is proportional to the ordinate of that point is (a) `( b ) (c) x=y(a-blogx)( d )` (e) (f) `( g ) (h)logx=b (i) y^(( j )2( k ))( l )+a (m)` (n) (o) `( p ) (q) (r) x^(( s )2( t ))( u )=y(a-blogy)( v )` (w) (d) None of these

Options

A. `x=y(a-blogy)`
B. `log_(x)=by^(2)+a`
C. `x^(2)=y(a-blogy)`
D. None of these

Answer

A

Explanation

Let the equation of the curve be `y=f(x)`

It is given that `OT propto y`
or OT=by
or OM-TM=by
or `x-y/(dy//dx)=by` [`therefore`=Length of the sub-tangent]
or `x-y(dx)/(dy)=by`
or `(dx)/(dy)-x/y=-b`
It is linear differential equation.
Its solution is `x/y=-blogy+a`
or `x=y(a-blogy)`