Tangent Points for Common Tangent to Two Ellipses

This is somewhat similar to my other question here.

Consider the two ellipses given by the equations

$$\frac{x^2}{2^2} + \frac{(y-1)^2}{1^2} = 1$$

and

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

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.)

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 $$x_2=-\frac{\sqrt{35}}{6}$$

• This looks really nice! Can you please extend your explanations, since I don't understand them to be honest? :-) – Svend Tveskæg Apr 7 '17 at 18:13
• I have added some extra explanations, but have not included all the algebra. Please let me know if there is anything you don't understand. – David Quinn Apr 7 '17 at 18:55
• You have accepted another answer, so I don't see why you need me to elaborate further on mine. But since you ask, the first quadratic equation is derived from multiplying out and simplifying $$\frac{x^2}{4}+(mx+c-1)^2=1$$ which is quicker for you to do yourself on paper than it is for me to type out in MathJax. :) – David Quinn Apr 14 '17 at 21:15
• No the method is independent of the actual ellipse equations. When a line touches a curve there must be a double root... – David Quinn Apr 14 '17 at 21:42
• It's the solution of $Ax^2+Bx+C=0$ when $B^2-4AC=0$ – David Quinn Apr 15 '17 at 5:29

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 \$$

• It’s worth pointing out that this solution works because the ellipses are homothetic. – amd Apr 7 '17 at 19:29
• @SvendTveskæg: I had to edit my answer. – zoli Apr 8 '17 at 12:06
• @SvendTveskæg: Thank you, but I had one more mistake. Take a look at the equations of the blue tangents. – zoli Apr 9 '17 at 7:16
• $OBD$ and $O'CD$ are similar because their angles equal. At $O$ and $C$ they have right angles and $D$ is a common angle. The third angles have to be equal because the sum of the angles is $180^{\circ}$. – zoli Apr 11 '17 at 20:35
• @SvendTveskæg: Not a stupid question. I was wrong. Of course the right result is $\pm \sqrt{3}/2$. – zoli Apr 12 '17 at 10:24