# Minimizing area of ellipse

Find minimum area of an ellipse that can pack three unit circles such that all three touch the ellipse internally: I took a point H as shown in the diagram and used the fact that the radius of the circle is 1, and that the circle touches the ellipse at point H. I am getting four equations for five unknowns, which means I can derive a relation between a and b of ellipse and use calculus to minimize the area. But those equations are tedious to solve and even after hours, I am not able to solve them. Is there any easier way to solve this? Here are the equations which I've got: • Do circles A and C touch the ellipse each at two points? Dec 30, 2018 at 16:18
• Why not to add the equations to the post ? Dec 30, 2018 at 16:22
• Is the angle $\theta$ given, or does it have to be chosen appropriately? Dec 30, 2018 at 16:28
• No it has to be chosen appropriately Dec 30, 2018 at 16:36
• @Claude Sorry, I am new to this site and I don't know mathjax. I can add a picture though. Dec 30, 2018 at 16:43

If we choose coordinates as in your sketch: $$A=(-2\sin\theta,0);\quad B=(0,-2\cos\theta);\quad C=(2\sin\theta,0);$$ then the tangent ellipse must be centered at $$(0,0)$$ and have $$b=1+2\cos\theta$$ as the $$y$$ semi-axis. Hence its equation is: $${x^2\over a^2}+{y^2\over(1+2\cos\theta)^2}=1,$$ where the unknown semi-axis $$a$$ must be determined so that the ellipse touches circle $$C$$. To find $$a$$ we can couple the equation of the ellipse to that of circle $$C$$: $$(x-2\sin\theta)^2+y^2=1$$ and plug then $$y^2=1-(x-2\sin\theta)^2$$ into the ellipse equation. The resulting quadratic equation in $$x$$ must have vanishing discriminant, which leads to: $$a={1+2\cos\theta\over\sqrt{\cos\theta}}.$$ Knowing both semi-axes $$a$$ and $$b$$ as a function of $$\theta$$ you can then find by yourself the minimum value of the area, which occurs for $$\cos\theta=1/6$$.

EDIT.

The above method to find $$a$$ works as long as tangency points have $$y\ne0$$, that is for $$\theta>\theta_0$$, with $$\theta_0\approx69.65°$$. For smaller values of $$\theta$$ you get simply $$a=1+2\sin\theta$$, but you can check that those ellipses have larger area.

• Did you use a software or something to find that expression for semi major axis a? I am still not able to find that expression. The equations are difficult to solve by hand. Dec 30, 2018 at 19:33
• I did it by hand. Remember you don't have to solve the equation for $x$, but just $\Delta=0$. Dec 30, 2018 at 20:18
• @Harsh in case you don't know this, the reason we set $\Delta=0$ (the discriminant) is because the circle is tangent to the ellipse, hence the quadratic equation must have precisely one real root. Dec 30, 2018 at 23:58

Let me try to help with the discriminant.

Start with $$\frac{x^2}{a^2}+ \frac{y^2}{(1+2\cosθ)^2}=1$$

Put $$y^2=1−(x−2 \sinθ)^2$$

So the equation becomes $$\frac{x^2}{a^2}+ \frac{1−(x−2 \sinθ)^2}{(1+2\cosθ)^2}=1$$

Compare it with the standard quadratic equation $$px^2 + qx + r = 0$$. If the discriminant is zero, then the roots are equal and $$q^2 = 4pr$$

Here, $$p = \frac{1}{a^2} - \frac{1}{(1+2 \cos θ)^2}$$ $$q = \frac{4 \sin θ}{(1+2 \cos θ)^2}$$ $$r = \frac{1- 4 \sin^2 θ}{(1+2 \cos θ)^2}-1$$

Now $$q^2 = 4pr \implies$$

$$\frac{16 \sin^2 θ}{(1+2 \cos θ)^4} = 4 \left(\frac{1}{a^2} - \frac{1}{(1+2 \cos θ)^2}\right) \left(\frac{1- 4 \sin^2 θ}{(1+2 \cos θ)^2}-1\right) (1)$$

Next, observe that $$1- 4 \sin^2 θ - (1+2 \cos θ)^2 = -4(1+ \cos θ)$$

Use this in (1) above:

$$\frac{16 \sin^2 θ}{(1+2 \cos θ)^4} = 4 \left(\frac{1}{a^2} - \frac{1}{(1+2 \cos θ)^2}\right) \left(\frac{-4(1+ \cos θ)}{(1+2 \cos θ)^2}\right) (2)$$

Next cancel $$16$$ from both LHS and RHS and write $$\sin^2 \theta = (1 - \cos \theta)(1 + \cos \theta)$$ on the LHS of (2) to obtain

$$\frac{(1 - \cos \theta)(1 + \cos \theta)}{(1+2 \cos θ)^4} = \left(\frac{1}{a^2} - \frac{1}{(1+2 \cos θ)^2}\right) \left(\frac{-(1+ \cos θ)}{(1+2 \cos θ)^2}\right) (3)$$

Next cancel $$(1 + \cos \theta)$$ from both LHS and RHS of (3) and rearrange to obtain

$$\frac{1}{a^2} = \frac{\cos \theta}{(1+2 \cos \theta)^2}$$