A curve is such that the mid-point of the portion of the tangent intercepted between the point where the tangent is drawn and the point where the tangent meets the y-axis lies on the line `y=xdot` If the curve passes through `(1,0),` then the curve is (a) `( b ) (c)2y=( d ) x^(( e )2( f ))( g )-x (h)` (i) (b) `( j ) (k) y=( l ) x^(( m )2( n ))( o )-x (p)` (q) (c) `( d ) (e) y=x-( f ) x^(( g )2( h ))( i ) (j)` (k) (d) `( l ) (m) y=2(( n ) (o) x-( p ) x^(( q )2( r ))( s ) (t))( u )` (v)

Options

A. `2y=x^(2)-x`
B. `y=x^(2)-x`
C. `y=x-x^(2)`
D. `y=2(x-x^(2))`

Answer

C

Explanation

The point on y-axis is `(0,y-x(dy)/(dx))`
According, to the given condition,
`x/2=y-x/2(dy)/(dx)`
or `(dy)/(dx) = 2(y/x)-1`
Putting `y/x=v`, we get `x(dv)/(dx)=v-1`
or `"ln"|y/x-1|="ln"|x|+c`
or `1-y/x=x` [as y(1)=0]