find the locus of the vertices of equilateral triangle circumscribing the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Question

My try : I am confused in this question , I have only tried questions who says to find the locus of a point but here we have to find locus of three points and there are only two relation , seems pretty odd and new to me. Need your help in this.

Answer 1

It should be an isoptic curve for an ellipse:

$$(x^2+y^2-a^2-b^2)^2 \tan^2 \alpha=4(a^2 y^2+b^2 x^2-a^2 b^2)$$

In this case, it should be the outer curve of the quartic with $\tan^2 \alpha=3$.

Answer 2

We want to find 3 points on the ellipse such that the tangent lines to those points intersect at the vertices of an equilateral triangle. So, given some point on the ellipse, what are other points that can be used to form an equilateral triangle?

Parameterize the ellipse with :

$$x=a\cos{\theta}$$ $$y=b\sin{\theta}$$

Select a point on the ellipse by fixing a value for $\theta$, e.g. $\theta=\theta_1$

at that point, $\frac{dx}{d\theta}=-a\sin{\theta}$ and $\frac{dy}{d\theta}=b\cos{\theta}$

So $\frac{dy}{dx}=\frac{-b\cos{\theta}}{a\sin{\theta}}$. Given a point on the line and a slope for the line, an equation of the line can be found.

Pick a second $\theta_2$ thus creating a new line.

Where do they intersect?

At what angle do they intersect?

Because it is to form a vertex of a triangle, we require those lines to intersect at a $60^\circ$ degre angle which has a cosine of $\frac{1}{2}$.

This requirement restricts the possible values of $\theta_2$.

This also restricts the values of a $\theta_3$.

The various values of theta imply the intersection of lines at the vertices of an equlateral triangle.

This should allow you to parameterize the location of vertices of an equlateral triangle given a single $\theta_1$ thus establishing the locus of points.