## Locus of Midpoint of tangent to Hypernola

Suppose a tangent is drawn to an hyperbola. The hyperbola intersects the two axes to form a line of finite length. What is the equation of the midpoint of this line?
A general point on the hyperbola is given in parametric coordinates as
$(acosht, b sinht)$
and the gradient of the hyperbola at this point is
$\frac{dy}{dx}= \frac{dy/dt}{dx/dt}= \frac{bcosht}{asinht}$
.
The
$y$
intercept - using
$y=mx+c$
- is the solution
$c$
to
$bsinht= \frac{bcoshtt}{asinht} \times acosht+c \rightarrow c=bsinht-\frac{bcosht}{asinht} \times acohst=\frac{bsinh^2t-bcosh^2t}{sinht}=-\frac{b}{sinht}$

The equation of the tangent is
$y= \frac{bcosht}{asinht}x-\frac{b}{sinht}$
.
The
$x$
intercept is the solution to
$0= \frac{bcosht}{asinht}c-\frac{b}{sinht} \rightarrow x= \frac{b/sinht}{bcosht/asinht}=\frac{a}{cosht}$

The
$x$
intercept is
$\frac{b}{sinht}$
.
The coordinates of the intercepts are
$(\frac{a}{cosht},0)$
and
$(0,\frac{b}{sinht})$
and the coordinates of the midpoints are
$(\frac{a}{2cosht}, \frac{b}{2sinht} )$
.
We get
$x=\frac{a}{2cosht} \rightarrow cosht=\frac{a}{2x}$
and
$y=\frac{b}{2sinht} \rightarrow sint= \frac{b}{2y}$
.
Then
$cosh^2-sinh^2t=1=\frac{a^2}{4x^2}-\frac{b^2}{4y^2}$
.