Finding angle between two tangents to the ellipse. How do I find the acute angle between two tangents which can be drawn from the point $(1,1)$ to the ellipse $4x^2+9y^2=1$ ? (example)
or simply say, how do I find the angle between two tangents which can be drawn from the point $(x,y)$ to the ellipse $\frac{x^2}{a} + \frac{y^2}{b} = 1$ ?
A graphical representation would be much appreciated, thanks !
 A: We are going to solve this problem in the general case by considering the ellipse as a parametric curve.
The ellipse defined by the equation
$$\left(\frac xa\right)^2+\left(\frac yb\right)^2=1$$
is represented parametrically as
$$z:(a\cos t,b\sin t)$$
and the derivative vector of the ellipse at some parameter $t$ is
$$z':(-a\sin t,b\cos t)$$
Now given a point $p=(p_x,p_y)$ and some point on the ellipse $z(t)$, the line between $p$ and $z(t)$ is a tangent to the ellipse if it is parallel to the derivative vector. In other words, the "planar cross product" $(a,b)\times(c,d)=ad-bc$ of $z(t)-p$ and $z'(t)$ has to be zero:
$$(a\cos t-p_x)(b\cos t)-(b\sin t-p_y)(-a\sin t)=0$$
$$ab\cos^2t-p_xb\cos t+ab\sin^2t-p_ya\sin t=0$$
$$ab=p_xb\cos t+p_ya\sin t$$
$$\frac{p_x}a\cos t+\frac{p_y}b\sin t=1$$
This last equation has two solutions for $t$ if $p$ is outside (not on) the ellipse; call these solutions $t_1$ and $t_2$. These correspond to two points on the ellipse $z_1$ and $z_2$, and $pz_1$ and $pz_2$ are the two tangents to the ellipse from $p$. Then the angle between the tangents can be computed from their dot product:
$$\theta=\cos^{-1}\frac{(z_1-p)\cdot(z_2-p)}{|z_1-p||z_2-p|}$$

For the given ellipse of $4x^2+9y^2=1$ ($a=\frac12,b=\frac13$) and $p=(1,1)$, $t=2\tan^{-1}(1\pm\frac2{\sqrt3})$. The slopes of the two tangents are $\frac49(3\pm\sqrt3)$, and the angle between them works out to be
$$\cos^{-1}\frac{59}{\sqrt{5209}}$$
or approximately 35.2 degrees. Below is a picture of this ellipse and the tangents to it.

