Question

A tangent line is drawn to the hyperbola `xy=c` at a point P, how do you show that the midpoint of the line segment cut from this tangent line by the coordinate axes is P?

Answer 1

We have `xy =c`, so differentiating simplicity (and using the product rule) gives:

`(x)(d/dxy) + (d/dxx)(y) = 0`
`:. xdy/dx + y = 0`
`:. dy/dx = -y/x`

Let us suppose that P has x-coordinates `t`, then `xy =c => y=c/x`, so P has coordinates `(t, c/t)`

So the gradient of the tangent at P is given by `dy/dx|_(x=t)`,
when `x=t => dy/dx=-y/x = -(c/t)/t = -c/t^2`.

The tangent passes through `(t, c/t)` and has gradient `m=-c/t^2`, so using `y-y_1=m(x-x_1)` the tangent has equation:
`y - c/t=-c/t^2(x - t)`
`:. t^2y - ct=-c(x - t)`
`:. t^2y - ct=-cx + ct`
`:. t^2y +cx = 2ct`

Now Let's find the midpoint of the tangent line as its passes through the axis.

The tangent cuts the `x`-axis when `y=0 => 0+cx=2ct`
`:. x=2t`

The tangent cuts the `y`-axis when `x=0 => t^2y+0=2ct`
`:. t(ty-2c)=0`
This yields two solutions:

1) Either, `t=0` which corresponds to a vertical asymptote `(0,oo)` which we can dismiss as a valid solution
2) Or `ty-2c=0 => y=(2c)/t`

So the tangent touches the axis at `(2t,0)` and `(0,(2c)/t)`

So the mid-pint of these two coordinates is:
`M=((2t+0)/2, (0+(2c)/t)/2 ) = (t,c/t)`, which is the same coordinate as P QED