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 (a) `( b ) (c)2(( d ) (e) (f) x^(( g )2( h ))( i )-( j ) y^(( k )2( l ))( m ) (n))=3x (o)` (p) (b) `( q ) (r)2(( s ) (t) (u) x^(( v )2( w ))( x )-( y ) y^(( z )2( a a ))( b b ) (cc))=6y (dd)` (ee) (c) `( d ) (e) x(( f ) (g) (h) x^(( i )2( j ))( k )-( l ) y^(( m )2( n ))( o ) (p))=6( q )` (r) (d) `( s ) (t) x(( u ) (v) (w) x^(( x )2( y ))( z )+( a a ) y^(( b b )2( c c ))( d d ) (ee))=10 (ff)` (gg)

Options

A. `2(x^(2)-y^(2))=3x`
B. `2(x^(2)-y^(2))=6y`
C. `x(x^(2)-y^(2))=6`
D. `x(x^(2)+y^(2))=10`

Answer

A

Explanation

`(dy)/(dx) = (x^(2)+y^(2))/(2xy)`………………(1)
Put `y=vx`, i.e., `(dy)/(dx) =v+x(dv)/(dx)`
Thus, equation (1) transforms to
`v+x(dv)/(dx) = (x^(2)v^(2)x^(2))/(2xvx) -v=(1+v^(2))/(2v)`
or `x(dv)/(dx)= (1+v^(2))/(2v) -v=(1-v^(2))/(2v)`
or `(2vdv)/(1-v^(2))=(dx)/x`
or `logx+log(1-v^(2))=logC`
or `logx+log(1-v^(2))=logC`
or `x(1-v^(2))=C`
or `x(1-y^(2)/x^(2))=C`
or `x^(2)-y^(2)=Cx`
If passes through (2,1).
`therefore 4-1=2c` or `C=3/2`
`therefore x^(2)-y^(2)=3/2x` or `2(x^(2)-y^(2))=3x`