Q 8 - Calculus

Question

Three noncollinear points determine a unique circle. Do three noncollinear points determine a unique ellipse? If so, explain why. If not, provide three noncollinear points that are on two distinct ellipses. 

Step-by-Step Solution

Answer

No, three non collinear points does not determine a unique ellipse.

For example :-

Graph the following two ellipses :-

$\frac{x^2}{4}+\frac{y^2}{16}=1 and \frac{x^2}{16}+\frac{{\left(y+2\right)}^2}{4}=1$ as following :-

There are three non collinear points $\left(0,-4\right),\left(-1.996,-0.267\right) and \left(1.996,-0.267\right)$ and these points are on two distinct ellipses. So these three points $\left(0,-4\right),\left(-1.996,-0.267\right) and \left(1.996,-0.267\right)$ are required points show that three non collinear points does not determine a unique ellipse as these are lies on two different ellipses.

Step 1. Given Information

We have to check that the three noncollinear points determine a unique ellipse or not.

If they forming a unique ellipse, then we have to explain it.

If not then we have to provide three noncollinear points that lies on two distinct ellipses.

Step 2. Explanation and example

We know that three noncollinear points determine a unique circle.

But this is not true in the case of ellipses. Three non collinear points are not enough to determine a unique ellipse. To determine a unique ellipse, we at least need five noncollinear points.

For example :-

Graph the following two ellipses :-

$\frac{x^2}{4}+\frac{y^2}{16}=1 and \frac{x^2}{16}+\frac{{\left(y+2\right)}^2}{4}=1$.

The graph of these two ellipses are as following :-

We can see that these ellipses have three points in common $\left(0,-4\right),\left(-1.996,-0.267\right) and \left(1.996,-0.267\right)$.

These three points are non collinear points.

That is these three non collinear points does not determine a unique ellipse as two different ellipses passes through these points.

So we can conclude that three non collinear points does not determine a unique ellipse.

Hence proved.