Question

An ellipse is described by using an endless string which is passed over two pins.

If the axes are $6cm$ and $4cm$, the length of the string and the distance between the pins are ____________

Answer 1

Step 1: Find the equation of the ellipse

Let us first of all derive the expression of the ellipse formed. The general expression for a standard ellipse is given by $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Given the axes are $6cm$ and $4cm$

Therefore $a=6,b=4$

Therefore the equation of the ellipse is

$\frac{x^2}{6^2}+\frac{y^2}{4^2}=1$

Step 2: Find the length of the string

Let's now determine the length of string required to create this ellipse. This length can be computed as follows and is equal to the ellipse's circumference:

Let the length of the string be $L$, then we have

$$\begin{gathered} L=\pi \left(a+b\right) \\ =\left(3.14\right)\left(6+4\right) \\ =31.4cm \end{gathered}$$

Therefore, the length of the string comes out to be $31.4cm$

Step 3: Find the distance between the two pin

We will now determine the separation between the two pins.

This has the value of $\left(2ae\right)$, where $"e"$ is the ellipse's eccentricity, which may be determined using the method shown below:

Where eccentricity$\left(e\right)$ of an ellipse is given by $e=\sqrt{1-\frac{b^2}{a^2}}$

$$\begin{gathered} \therefore e=\sqrt{1-\frac{b^2}{a^2}} \\ \Rightarrow e=\sqrt{1-\frac{16}{36}} \\ \Rightarrow e=\frac{\sqrt{20}}{6} \\ \Rightarrow e=\frac{\sqrt{5}}{3} \end{gathered}$$

Therefore, the length between the two foci (say, $d$) can be calculated as follows:

$$\begin{gathered} \Rightarrow d=2ae \\ \Rightarrow d=2\times 6\times \frac{\sqrt{5}}{3} \\ \Rightarrow d=4\sqrt{5}cm \end{gathered}$$

Therefore the ellipse can be constructed as the following figure

The pins that are pinned down at $F_1$ and $F_2$, as well as the distinctive string-encircled ellipse, are easily visible.
Therefore, the distance between the two pins is therefore equal to $4\sqrt{5}cm$, and the required length of the string is $31.4cm$.