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

Options

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

Answer

A

Explanation

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.`