Home
Class 12
MATHS
A curve passes through the point (1,pi/6...

A curve passes through the point `(1,pi/6)` . Let the slope of the curve at each point `(x , y)` be `y/x+sec(y/x),x > 0.` Then the equation of the curve is (a) `( b ) (c)sin(( d ) (e) (f) y/( g ) x (h) (i) (j))=logx+( k )1/( l )2( m ) (n) (o)` (p) (q) `( r ) (s)"c o s e c"(( t ) (u) (v) y/( w ) x (x) (y) (z))=logx+2( a a )` (bb) (cc) `( d d ) (ee)sec(( f f ) (gg) (hh)(( i i )2y)/( j j ) x (kk) (ll) (mm))=logx+2( n n )` (oo) (pp) `( q q ) (rr)cos(( s s ) (tt) (uu)(( v v )2y)/( w w ) x (xx) (yy) (zz))=logx+( a a a )1/( b b b )2( c c c ) (ddd) (eee)` (fff)

A

(a) `sin(y/x)=log x+ 1/2`

B

(b) cosec `(y/x)=log x +2`

C

(c) `sec((2y)/x)=log x +2`

D

(d) `cos((2y)/x)=log x +1/2`

Text Solution

Verified by Experts

The correct Answer is:
(a)
PLAN To solve homogeneous differential equation, i.e. substitute
`y/x=V`
`therefore y=vx rArr dy/dx=v+s(dv)/dx`
Here, slope of the curve at (x,y) is
`dy/dx=y/x+sec(y/x)`
Put `y/x= v`
`therefore v+x(dv)/dx=v+sec(v) rArr x(dv)/dx=sex(v)`
`rArr int (dv)/(secv)=intdx/x rArr intcos v dv=int dx/x`
`rArr sinv=logx +log c rArr sin (y/x)=log(cx)`
As it passes through `(1,pi/6) rArr sin (pi/6) =- log c `
`rArr logc=1/2`
`therefore sin (y/x)=log x+1/2 `
Promotional Banner

Similar Questions

Explore conceptually related problems

If a curve passes through the point (1, -2) and has slope of the tangent at any point (x,y) on it as (x^2-2y)/x , then the curve also passes through the point

The slope of the tangent at (x , y) to a curve passing through a point (2,1) is (x^2+y^2)/(2x y) , then the equation of the curve is

The slope of the tangent at (x , y) to a curve passing through (1,pi/4) is given by y/x-cos^2(y/x), then the equation of the curve is

The differential equation of the curve x/(c-1)+y/(c+1)=1 is

The circumcenter of the triangle formed by the line y=x ,y=2x , and y=3x+4 is (a) ( b ) (c)(( d ) (e)6,8( f ))( g ) (h) (b) ( i ) (j)(( k ) (l)6,8( m ))( n ) (o) (c) ( d ) (e)(( f ) (g)3,4( h ))( i ) (j) (d) ( k ) (l)(( m ) (n)-3,-4( o ))( p ) (q)

The slope at any point of the curve y = f(x) is given by (dy)/(dx)= 2x it passes through (1,-1) then the equation of the curve is :

Find the equation of a curve passing through the point (-2,3), given that the slope of the tangent to the curve at any point (x,y) is (2x)/(y^(2)) .

Find the equation of the curve passing through (1,0) and which has slope 1+(y)/(x) at (x,y)

The gradient (siope) of a curve at any point (x,y) is (x^(2)-4)/(x^(2)) . If the curve passes through the point (2 , 7) , then the equation of the curve is ,

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