A curve `C` has the property that if the tangent drawn at any point `P` on `C` meets the co-ordinate axis at `A` and `B` , then `P` is the mid-point of `A Bdot` The curve passes through the point (1,1). Determine the equation of the curve.

Answer

`x^(2)+y^(2)=ky`

Explanation

Equation of the tangent at point (x,y) on the curve is
`Y-y-(dy)/(dx)(X-x)1`
This meet axis is `A(x-y(dx)/(dy),0)` and `B(0,y-x(dy)/(dx))`.
Midpoint of AB is `(1/2x-y(dx)/(dy),0)` and `B(0,y-x(dy)/(dx))`
We are given `1/2(x-y(dx)/(dy))=x`and `(dy)/(dx) =-(dx)/x`
Integrating both sides, we get `int(dy)/(y) = - int(dy)/(y)`
or `logy=-logx+c`
Put `x=1, y=1`. Then `log 1-log1=c` or =0.
`therefore logy+logx=0` or `logyx=0`
or `yx=e^(0)=1` which is a rectangular hyperbola.