Tangent to a curve intercepts the y-axis...

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

`y.(dy)/(dx)-x((dy)/(dx))^(2)=1`

`x(d^(2)y)/(dx^(2))+((dy)/(dx))^(2)=1`

`y.(dx)/(dy)+x = 1`

None of these

Text Solution

Verified by Experts

The correct Answer is:

Equation of tangent at the point
`R(x, f(x)) "is Y" - f(x) = f'(x) (X -x)`
Coordinates of point P are (0, f(x) - x f'(x))
The slope of the prependicular line through 'P' is
`(f(x)-xf'(x))/(-1)=-(1)/(f'(x))`
`rArr" "y(dy)/(dx)-x((dy)/(dx))^(2)=1` is differential equation.

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