Find the coordinates of the point on the curve `y=x-(4)/(x)`, where the tangent is parallel to the line `y=2x`.

Explanation

Let `P(x_(1),y_(1))` be the required point on the curve `y=x-(4)/(x)`.
Differentiating `y=x-(4)/(x)` w.r.t. x, we get,
`(dy)/(dx)=(d)/(dx)(x-(4)/(x))=1-4(-(1)/(x^(2)))=1+(4)/(x^(2))`
`:.` slope of the tangent at `(x_(1),y_(1))`
`=((dy)/(dx))_("at"(x_(1),y_(1)))=1+(4)/(x_(1)""^(2))`
Since this tangent is parallel to the line `y=2x` whose slope is 2, slope of the tangent = 2
`:.1+(4)/(x_(1)""^(2))=2" ":.(4)/(x_(1)""^(2))=1`
`:.x_(1)""^(2)=4" ":.x_(1)=+-2`
Since `(x_(1),y_(1))` lies on the curve `y=x-(4)/(x)`,
`y_(1)=x_(1)-(4)/(x_(1))`
If `x_(1)=2, y_(1)=2-(4)/(2)=2-2=0`
If `x_(1)=-2,y_(1)=-2-(4)/((-2))=-2+2=0`
Hence, the coordinates of the required points are (2, 0) and (-2, 0).