## Locus of Midpoint of Tangent to Ellipse

Suppose a tangent is drawn to an ellipse. The ellipse 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 ellipse is given in parametric coordinates as
$(acost, b sint)$
and the gradient of the ellipse at this point is
$\frac{dy}{dx}= \frac{dy/dt}{dx/dt}=- \frac{bcost}{asint}$
.
The
$y$
intercept - using
$y=mx+c$
- is the solution
$c$
to
$bsint= - \frac{boost}{asint} \times acost+c \rightarrow c=bsint+\frac{bcost}{asint} \times acost=\frac{bsin^2t+bcos^2t}{sint}=\frac{b}{sint}$

The equation of the tangent is
$y=- \frac{bcost}{asint}c+\frac{b}{sint}$
.
The
$x$
intercept is the solution to
$0=- \frac{bcost}{asint}c+\frac{b}{sint} \rightarrow x= \frac{b/sint}{bcost/asint}=\frac{a}{cost}$

The
$x$
intercept is
$\frac{b}{sint}$
.
The coordinates of the intercepts are
$(\frac{a}{cost},0)$
and
$(0,\frac{b}{sint})$
and the coordinates of the midpoints is
$(\frac{a}{2cost}, \frac{b}{2sint} )$
.
We get
$x=\frac{a}{2cost} \rightarrow cost=\frac{a}{2x}$
and
$y=\frac{b}{2sint} \rightarrow sint= \frac{b}{2y}$
.
Then
$cos^2+sin^2t=1=\frac{a^2}{4x^2}+\frac{b^2}{4y^2}$
.