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.
 A: Using implicit differentiation, you can find the slope of a line tangent to the ellipse at a point $(a,b)$ on the ellipse. Taking the derivative,
$$
18x+8y\frac{dy}{dx}=0\implies \frac{dy}{dx}=\frac{-9x}{4y}.
$$
So the slope of a tangent line at a point $(a,b)$ is $\frac{-9a}{4b}$, so the equation of such is line is
$$
y-b=\frac{-9a}{4b}(x-a)
$$
which is equivalent to
$$
9ax+4by=9a^2+4b^2=36.
$$
Since this line must also pass through $(0,6)$, plugging in you find $24b=36$, or $b=3/2$. Substituting back into the original equation yields $a=\pm\sqrt{3}$, so the two points of tangency are $(\pm\sqrt{3},3/2)$.
A: A standard way to solve the problem is to consider the generic line passing through $(0,6)$ which has equation
$$
y-6=mx,
$$
then make a substitution in the ellipse equation and impose that the resulting one variable quadratic equation has a double root. This will give you the values $m$ of the tangent lines.
A: We use a linear transformation. Go from $(3x)^2+(2y)^2=6^2$ to $u^2+v^2=1$ by using a change of coordinates with $u=\frac{1}{2}x$ and $v=\frac{1}{3}y$. Tangencies and intersections are preserved, so there are two lines tangent to the unit circle intersecting the point $(0,2)$ in the $uv$-plane; they will be symmetric across the $v$-axis so finding one is enough. Make a right triangle with one vertex at the origin $O$, one at the point of tangency $T$ (say on the right side), and one at $P=(0,2)$, with $\phi=\angle POT$; then trigonometry dictates $\cos\phi=1/2\implies\phi=\pi/3$, hence the points in the $uv$-plane are
$$(\cos(\pi/2-\phi),\sin(\pi/2-\phi))=\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right),\text{ and}$$
$$(-\cos(\pi/2-\phi),\sin(\pi/2-\phi))=\left(-\frac{\sqrt{3}}{2},\frac{1}{2}\right).$$
Transform these points back into the $xy$-plane using $x=2u,y=3v$ and obtain
$(\pm\sqrt{3},3/2).$
A: 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 $\left(\pm\sqrt 3,\dfrac32\right)$.
