# 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?

Nov 15, 2016

We have $x y = c$, so differentiating simplicity (and using the product rule) gives:

$\left(x\right) \left(\frac{d}{\mathrm{dx}} y\right) + \left(\frac{d}{\mathrm{dx}} x\right) \left(y\right) = 0$
$\therefore x \frac{\mathrm{dy}}{\mathrm{dx}} + y = 0$
$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x}$

Let us suppose that P has x-coordinates $t$, then $x y = c \implies y = \frac{c}{x}$, so P has coordinates $\left(t , \frac{c}{t}\right)$

So the gradient of the tangent at P is given by $\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{x = t}$,
when $x = t \implies \frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{y}{x} = - \frac{\frac{c}{t}}{t} = - \frac{c}{t} ^ 2$.

The tangent passes through $\left(t , \frac{c}{t}\right)$ and has gradient $m = - \frac{c}{t} ^ 2$, so using $y - {y}_{1} = m \left(x - {x}_{1}\right)$ the tangent has equation:
$y - \frac{c}{t} = - \frac{c}{t} ^ 2 \left(x - t\right)$
$\therefore {t}^{2} y - c t = - c \left(x - t\right)$
$\therefore {t}^{2} y - c t = - c x + c t$
$\therefore {t}^{2} y + c x = 2 c t$

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

The tangent cuts the $x$-axis when $y = 0 \implies 0 + c x = 2 c t$
$\therefore x = 2 t$

The tangent cuts the $y$-axis when $x = 0 \implies {t}^{2} y + 0 = 2 c t$
$\therefore t \left(t y - 2 c\right) = 0$
This yields two solutions:

1) Either, $t = 0$ which corresponds to a vertical asymptote $\left(0 , \infty\right)$ which we can dismiss as a valid solution
2) Or $t y - 2 c = 0 \implies y = \frac{2 c}{t}$

So the tangent touches the axis at $\left(2 t , 0\right)$ and $\left(0 , \frac{2 c}{t}\right)$

So the mid-pint of these two coordinates is:
$M = \left(\frac{2 t + 0}{2} , \frac{0 + \frac{2 c}{t}}{2}\right) = \left(t , \frac{c}{t}\right)$, which is the same coordinate as P QED