How to find points of tangency on an ellipse? The problem I have to solve is:
    If tangent lines to ellipse $9x^2+4y^2=36$ intersect the y-axis at point $(0,6)$, find the points of tangency.

Yet another method: converting the equation of the ellipse into the form
$$\frac{x^2}{4}+\frac{y^2}{9}=1$$
and by exploiting the identity
$$\left(\frac{1-t^2}{1+t^2}\right)^2+\left(\frac{2t}{1+t^2}\right)^2=1$$
we obtain the parametrization
$$\begin{align*}x&=2\frac{1-t^2}{1+t^2}\\y&=\frac{6t}{1+t^2}\end{align*}$$
From this parametrization, we derive the equation of the tangent line:
$$y=\frac{3(t^2-1)}{4t} \left(x-2\frac{1-t^2}{1+t^2}\right)+\frac{6t}{1+t^2}$$
or, simplified,
$$y=\frac{3(t^2-1)}{4t}x+\frac{3\left(t^2+1\right)}{2t}$$
We ask that the $y$-intercept of the tangent line be $6$; equating the constant term of the linear equation to $6$ and rearranging yields
$$3 t^2-12 t+3=0$$
which has the roots $t=2\pm\sqrt{3}$; substituting these values of $t$ into the original parametric equations yields the tangency points $\left(\pm\sqrt 3,\dfrac32\right)$.
