The equation of a curve passing through (1,0) for which the product of the abscissa of a point `P` and the intercept made by a normal at `P` on the x-axis equal twice the square of the radius vector of the point `P` is (a) `( b ) (c) (d) x^(( e )2( f ))( g )+( h ) y^(( i )2( j ))( k )=( l ) x^(( m )4( n ))( o ) (p)` (q) (b) `( r ) (s) (t) x^(( u )2( v ))( w )+( x ) y^(( y )2( z ))( a a )=2( b b ) x^(( c c )4( d d ))( e e ) (ff)` (gg) (c) `( d ) (e) (f) x^(( g )2( h ))( i )+( j ) y^(( k )2( l ))( m )=4( n ) x^(( o )4( p ))( q ) (r)` (s) (d) None of these

Options

A. `x^(2)+y^(2)=x^(4)`
B. `x^(2)+y^(2)=2x^(4)`
C. `x^(2)+y^(2)=4x^(4)`
D. None of these

Answer

A

Explanation

Tangent at point P is `Y-y=-1/m(X-x)` where `m=(dy)/(dx)`
Let Y=0, Then `X=my+x`
According to question, `x(my+x)=2(x^(2)+y^(2))`
or `(dy)/(dx)=(x^(2)+2y^(2))/(xy)` (homogenous)
Putting `y=vx` we get
`v=x(dv)/(dx) = (1+2v^(2))/(v)`
or `x(dv)/(dx) = (1+2v^(2))/v`
`x(dv)/(dx) = (1+2v^(2))/(v) -v=(1+v^(2))/v`
or `int(vdv)/(1+v^(2))=int(dx)/x`
or `1/2log(1+v^(2))=logx+logc, c gt 0`
or `x^(2)+y^(2)=cx^(4)`
Also, it passes through (1,0). Then c=1.