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The slope of the tangent at (x , y) to a...

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)

Text Solution

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It is given that, the slope of tanget to the curve at point (x, y) is `(x^(2)+y^(2))/(2xy)`.
`therefore" "((dy)/(dx))_"(x,y)"=(x^(2)+y^(2))/(2xy)`
`rArr" "(dy)/(dx)=(1)/(2)((x)/(y)+(y)/(x))" "` …(i)
which is homogeneous differential equation.
Put `" "y=vx`
`rArr" "(dy)/(dx)=v+x(dv)/(dx)`
On substituting these values in Eq. (i), we get
`" "v+x(dv)/(dx)=(1)/(2)((1)/(v)+v)`
`rArr" "v+x(dv)/(dx)=(1)/(2)((1+v^(2))/(v))`
`rArr" "x(dv)/(dx)=(1+v^(2))/(2v)-v`
`rArr" "x(dv)/(dx)=(1+v^(2)-2v^(2))/(2v)`
`rArr" "x(dv)/(dx)=(1-v^(2))/(2v)`
`rArr" "(2v)/(1-v^(2))dv=(dx)/(x)`
On integrating both sides, we get
`" "int(2v)/(1-v^(2))dv=int(dx)/(x)`
Put `1-v^(2)-t` in LHS, we get
`" "-2vdv=dt`
`rArr" "-int(dt)/(t)=int(dx)/(x)`
`rArr" "-logt=logx+logC`
`" "-log(1-v^(2))=logx+logC`
`rArr" "-log(1-(y^(2))/(x^(2)))=logx+logC`
`rArr" "-log((x^(2)-y^(2))/(x^(2)))=logx+logC`
`rArr" "log((x^(2))/(x^(2)-y^(2)))=logx+logC`
`rArr" "(x^(2))/(x^(2)-y^(2))=Cx" "` ...(ii)
Since, the curve passes through the point (2,1).
`therefore" "((2)^(2))/((2)^(2)-(1)^(2))=C(2)rArrC=(2)/(3)`
So, the required solution is `2(x^(2)-y^(2))=3x.`
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