# An ellipse with major axis $4$ and minor axis $2$ touches both the coordinate axes. Locus of its Center and Focus is?

An ellipse with major axis $$4$$ and minor axis $$2$$ touches both the coordinate axes. Locus of its Center and Focus is?

My Approach: For locus of Center.

Since it is touching the coordinate axes, the coordinate axes will act as tangents making angle of $$90^{\circ}$$, so origin will lies on Director Circle.

Center of director circle will be same as center of Ellipse.

Let center of Ellipse be $$(h,k)$$, so the equation of director circle will be $$(x-h)^2+(y-k)^2=(\text{semi major axis})^2+ (\text{semi minor axis})^2,$$

$$i.e.\;\;\;(x-h)^2+(y-k)^2=(2)^2+ (1)^2.$$

Because it passes through the origin,

$$(0-h)^2+(0-k)^2=(2)^2+ (1)^2$$

$$\implies$$ $$h^2+k^2=5$$

$$\implies$$ locus of center of Ellipse is $$x^2+y^2=5$$.

For Locus of Focus:

I assumed focus as $$(x_1,y_1)$$ but i cannot proceed further. I know one property, that product of distance of tangents from Foci is constant and equal to square of semi-minor axis and lies on auxiliary circle, but that leads me nowhere.

Note There are many solutions available on internet, but they all did this with the same method of taking axis of ellipse parallel to coordinate axis.

I don't want to do using that method. I want it for slanted ellipse as shown in attached image.

• Please also see animation in my post here. Commented Apr 28, 2022 at 23:25

First, we try to determine the general form of a rotated ellipse in the first quadrant such that it remains tangent to the axes. We suppose it has the parametric equation

$$(x(t),y(t)) = (4 \cos u \cos t - 2 \sin u \sin t + h, 2 \cos u \sin t + 4 \sin u \cos t + k), \quad t \in [0,2\pi)$$ where $$(h,k)$$ is the center, and $$u$$ is the counterclockwise rotation angle of the ellipse relative to the coordinate axes. (Note I have modified the direction of rotation compared to the linked answer.)

Such an ellipse has horizontal tangent lines satisfying $$0 = \frac{dy}{dt} = 2 \cos u \cos t - 4 \sin u \sin t,$$ or $$t_{\text{crit}} = \arctan \frac{\cot u}{2}.$$ For these values of $$t_{\text{crit}}$$, we need to find $$k$$ such that $$y(t_{\text{crit}}) = 0$$, placing this ellipse so it is tangent to the $$x$$-axis; i.e, $$k = 2 \sqrt{\cos^2 u + 4 \sin^2 u}.$$ This gives, as a function of the angle of rotation $$u$$, the necessary vertical translation to make the ellipse tangent. A similar process using $$dx/dt$$ gives the necessary horizontal translation, which we show without proof: $$h = 2 \sqrt{4 \cos^2 u + \sin^2 u}.$$ Thus our ellipse is fully parametrized.

The locus of the center is simply $$(h,k)$$ as a function of $$u$$:

$$(h(u), k(u)) = \left(2 \sqrt{4 \cos^2 u + \sin^2 u}, 2 \sqrt{\cos^2 u + 4 \sin^2 u}\right).$$

A short computation of $$h^2 + k^2$$ shows that this locus is an arc of a circle, not the complete circle.

Where are the foci? We can first observe that they are located at some point along the line joining $$(x(0), y(0))$$ and $$(x(\pi), y(\pi))$$, so they have coordinates of the form $$(1-\lambda)(x(0), y(0)) + \lambda (x(\pi), y(\pi))$$ for $$\lambda = \frac{4 + 2 \sqrt{3}}{8} = \frac{2 + \sqrt{3}}{4}, \quad \text{and} \quad \lambda = \frac{2 - \sqrt{3}}{4}.$$ We again skip the calculation and show the result: $$(x_f(u), y_f(u)) = 2 \left(\sqrt{3} \cos u + \sqrt{4 \cos^2 u + \sin^2 u}, \sqrt{3} \sin u + \sqrt{\cos^2 u + 4 \sin^2 u} \right). \tag{1}$$ Note that this curve gives the locus of both foci, where $$u$$ and $$u + \pi$$ represent the location of each focus for a given rotation angle $$u$$.

All put together, we can visualize these loci in the following animation:

The conversion of the parametric formula to an implicit curve is tedious but not intractable; one would start with computing the square, then show that the square of the locus satisfies $$(x + y)(16 + xy) = 64 xy.$$

One approach to converting the locus is to note that we can write $$(x_f(u), y_f(u)) = \left(2 \sqrt{3} \cos u + \sqrt{(2 \sqrt{3} \cos u)^2 + 4}, 2 \sqrt{3} \sin u + \sqrt{(2 \sqrt{3} \sin u)^2 + 4} \right),$$ therefore $$x_f(u)$$ and $$y_f(u)$$ are roots of the quadratics $$x^2 - (4 \sqrt{3} \cos u) x - 4 = 0, \\ y^2 - (4 \sqrt{3} \sin u) y - 4 = 0,$$ or equivalently, $$48 \cos^2 u = \frac{(4-x^2)^2}{x^2}, \quad 48 \sin^2 u = \frac{(4-y^2)^2}{y^2}.$$ Thus $$48 = \frac{(4-x^2)^2}{x^2} + \frac{(4-y^2)^2}{y^2},$$ and the rest is an exercise in algebra.

• In which software did you do animation? Commented Jun 17, 2021 at 1:35
• @mathophile The animation was created in Mathematica version 12. Commented Jun 17, 2021 at 3:01

The method used in the other answer was pretty neat, but I found another approach using the property you mentioned.

Assume one of the foci to be $$(k,h)$$, and the angle made by the ellipse's major axis with the $$x$$-axis to be $$\alpha$$.

We know the distance between two foci to be $$2ae$$, where $$a=2$$ and $$e=\frac{\sqrt{3}}{2},$$ so the other foci will be:

$$\big(k+2\sqrt{3}\cos(\alpha),\;h+2\sqrt{3}\sin(\alpha)\big).$$

Using the property that you mentioned of ellipse that the product of the length of the perpendicular segments from the foci on any tangent to the ellipse is $$b^2$$, we get:

$$k(k+2\sqrt{3}\cos(\alpha))=h(h+2\sqrt{3}\sin(\alpha))=b^2=1,$$

as the perpendicular from two foci on tangent $$x=0$$ is $$k$$ and $$k+2\sqrt{3}\cos(\alpha)$$.

Thus we get $$\cos(\alpha)=\frac{1-k^2}{2k\sqrt3}$$ and $$\sin(\alpha)=\frac{1-h^2}{2h\sqrt3}.$$ On squaring and adding, we get $$h^2+k^2+\frac{1}{k^2}+\frac{1}{h^2}=16.$$ Replacing $$(k,h)$$ with $$(x,y)$$: $$x^2+y^2+\frac{1}{x^2}+\frac{1}{y^2}=16.$$