Tangent to a curve intercepts the y-axis at a point `Pdot` A line perpendicular to this tangent through `P` passes through another point (1,0). The differential equation of the curve is (a) `( b ) (c) y (d)(( e ) dy)/( f )(( g ) dx)( h ) (i)-x (j) (k)(( l ) (m) (n)(( o ) dy)/( p )(( q ) dx)( r ) (s) (t))^(( u )2( v ))( w )=1( x )` (y) (b) `( z ) (aa) (bb)(( c c ) x (dd) d^(( e e )2( f f ))( g g ))/( h h )(( i i ) d (jj) x^(( k k )2( l l ))( m m ))( n n ) (oo)+( p p ) (qq)(( r r ) (ss) (tt)(( u u ) dy)/( v v )(( w w ) dx)( x x ) (yy) (zz))^(( a a a )2( b b b ))( c c c )=0( d d d )` (eee) (c) `( d ) (e) y (f)(( g ) dx)/( h )(( i ) dy)( j ) (k)+x=1( l )` (m) (d) None of these

Options

A. `y(dy)/(dx) -x((dy)/(dx))^(2)=1`
B. `(xd^(2)y)/(dx^(2))+((dy)/(dx))^(2)=0`
C. `y(dy)/(dx)+x=1`
D. None of these

Answer

A

Explanation

The equation of the tangent at the point `R(x,f(x))` is `Y-f(x)=f^(')(x)(X-x)`
The coordinates of the point P are `(0,f(x)-xf^(')(x))`.
The slope of the perpendiular line through P is
`(f(x)-xf^(')(x))/-1 = -1/(f^(')x)`
or `f(x)f^(')(x)-x(f^(')(x))^(2)=1`
or `(ydx)/(dx) -x((dy)/(dx))^(2)=1`
Which is the required different equation to the curve at `y=f(x)`