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?

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