Given a formula for an ellipse, {x^2\over a^2} + {y^2 \over b^2} = 1 and a point Q(x_1,y_1) outside of the ellipse, to find the point of contact P(x, y) on the ellipse:
There's an intuitive solution and a solution using differentiation.
I'll show you the differentiation version and then show you it is equivalent to the intuitive one.
Here goes the differentiation version:
let u={x\over a}; v={y\over b},
then u^2 + v^2 = 1 and it is a formula for a circle of unit radius.
You can parametrize this and say that
u = x/a = \cos t, and
v = y/b = \sin t.
and you get x = a \cos t and y = b \sin t.
Thus P=(a \cos t, b \sin t)
The vector along the tangential line at P is parallel to {dP\over dt} = (-a \sin t, b \cos t).
Thus the cross product (Q-P) \times {dP\over dt} = 0
Thus
(x_1-a \cos t) b \cos t + (y_1- b \sin t) a \sin t = 0.
Now divide the both sides by ab and you get
(x_1/a - \cos t) \cos t + (y_1/b - \sin t) \sin t = 0,
or
{x_1\over a} \cos t + {y_1\over b} \sin t = 1, since \cos^2 t+\sin^2 t = 1
Now define D as
D^2 = u_1^2 + v_1^2, where u_1 = x_1/a and v_1 = y_1/b
and also define
\tan\theta = v_1/u_1 ; \cos\theta = u_1/D ; \sin\theta = v_1/D
and also define \cos\phi = 1/D
Then the equation is
u_1 \cos t + v_1 \sin t = 1,
or
D (\cos\theta \cos t + \sin\theta \sin t) = 1,
or \cos(\theta-t) = 1/D = \cos\phi.
so you get
\theta - t = \pm\phi,
or t = \theta \pm\phi, where \theta = \arctan(1/D) and \phi = \arccos(v_1/u_1).
and D=sqrt(u_1^2 + v_1^2), and u_1=x_1/a, v_1=y_1/b, where {x_1} and {y_1} are coordinates of Q.
then P is given as P = (a \cos (\theta\pm\phi), b \sin (\theta\pm\phi)).
In other words to get the parameter t where the line contacts P
draw a triangle with a right angle that has a base of x_1/a and height y_1/b.
Then the hypotenuse is D(>1 if Q is outside of the ellipse).
Draw a circle of unit radius with the lower end of the hypotenuse as its center.
cast a tangent line from the upper end of the hypotenuse to the circle.
The angle between the base of the triangle and the hypotenuse is \theta,
and the angles between the hypotenuse and the line joining the center of the circle and the contact points
are \pm\phi each.
and t is anyone of the sum of angles say \theta+\phi or \theta-\phi.
So you can see it is equivalent to the figure that you will get if you shrinked the ellipse by a horizontally
and by b vertically, and the coordinates of the Q accordingly, into say Q'(x_1/a, y_1/b),
then drew the tangent lines from Q' to the shrunk ellipse, now a unit circle.
The complicated mathematical operations I have performed up above is just to show the validity of this intuitional operation. so you know what to do now:
1. Shrink the coordinate system uniformly until the ellipse becomes a unit circle.
2. draw the tangent line from Q' to the resulting unit circle and get the contact points P'.
3. stretch the coordinate syst\em back to the original one(P' --> P).