Tangent Points for Common Tangent to Two Ellipses

Question

This is somewhat similar to my other question here.

Consider the two ellipses given by the equations

\begin{equation} \frac{x^2}{2^2} + \frac{(y-1)^2}{1^2} = 1 \end{equation}

and

\begin{equation} \frac{x^2}{1^2} + \frac{(y-4)^2}{(1/2)^2} = 1. \end{equation}

How do I find the coordinates for the two tangent points of their common tangent at the left side of the ellipses? (I hope the question makes sense.)

Answer 1

Let such a tangent line be $y=mx+c$

Solving simultaneously with the ellipses gives two quadratics in $x$ which are $$(4m^2+1)x^2+8mx(c-1)+4c^2-8c=0$$ and $$(4m^2+1)x^2+8mx(c-4)+4c^2-32c+63=0$$

Each of these must have double roots at the point of tangency and therefore the discriminant is zero, so this leads to two equations in $m$ and $c$ which are $$64m^2(c-1)^2=4(4m^2+1)(4c^2-8c)$$ and $$64m^2(c-4)^2=4(4m^2+1)(4c^2-32c+63)$$

Dividing these eliminates the $m$ terms and we get an equation for $c$ which is $$3c^2-30c+63=0\implies c=3,7$$

A diagram will confirm that the value $c=3$ gives us the tangents which pass between the ellipses, so we choose $c=7$ from which we get the gradient for the common tangent on the left as $$m=+\frac{\sqrt{35}}{2}$$

Using "$x=-\frac{B}{2A}$" the $x$-coordinates of the points of tangency which are $$x_1=-\frac{\sqrt{35}}{3}$$ and $$x_2=-\frac{\sqrt{35}}{6}$$

Answer 2

Let's introduce the following new variables: $$2u=x\ \text{ and }\ y-1=v.$$

With these new variables, we have

$$u^2+v^2=1\ \text{ and }\ u^2+(v-3)^2=\frac14$$

that is, we have two circles as shown in the figure below.

We have similar triangles and we can see that $OD=6$. Also, by the Pythagorean theorem $DC=\frac{\sqrt{35}}2$ and by the similarity of $OBD$ and $O'CD$: $OB=\frac6{\sqrt{35}}$. So, the slope of the red straight lien is $-\sqrt{35}$. Don't forget a about the other tangent, not shown, whose slope is $\sqrt{35}$.

The two tangent lines in the $u,v$ system are

$$v=-\sqrt{35}u+6\ \text{ and } \ v=\sqrt{35}u+6.$$

Returning to the $x,y$ coordinate system, we get

$$y=-\frac{\sqrt{35}}2x+7\ \text{ and } \ y=\frac{\sqrt{35}}2x+7.$$

EDIT

Unforgivable! I forgave the other pair of tangents:

After similar calculations we get the equations of the other pair of tangent lines.

$$y=-\frac{\sqrt3}{2}x+3\ \text{ and }y=\frac{\sqrt3}{2}x+3 \ $$