Find the eqution of the curve passing through the point (1,1), if the tangent drawn at any point P(x,y) on the curve meets the coordinate axes at A and B such that P is the mid point of AB.

Explanation

The below figure obtained by the given information

Let the coordinate of the point P is (x, y). It is given that, P is mid-point of AB.
So, the coordinates of points A and B are (2x, 0) and (0, 2y), respectively.
`therefore" "` Slope of `AB=(0-2y)/(2x-0)=-(y)/(x)`
Since, the segment AB is a tangent to the curve at P.
`therefore" "(dy)/(dx)=-(y)/(x)`
`rArr" "(dy)/(y)=-(dx)/(x)`
On integrating both sides, we get
`" "logy=-logx+logC`
`" "logy=log""(C)/(x)" "`...(i)
Since, the given curve passes through (1, 1).
`therefore" "log1=log""(C)/(1)`
`rArr" "0=logC`
`rArr" "c=1`
`therefore" "logy=log""(1)/(x)`
`rArr" "y=(1)/(x)`
`rArr" "xy=1`