Applications of Ellipse

Example 1: An arch is in the form of a semi-ellipse, and it is 48 feet wide at the base and has a height of 20 feet. How wide is the arch at the height of 10 feet above the base?


app-ellipse

Here the length of the major axis is given as 2a = 48\,\, \Rightarrow a = 24 and the height of the semi-ellipse is given as b = 20.

Now the standard equation of an ellipse is

\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1

Putting the values of a and b in the above equation of an ellipse, we have

\frac{{{x^2}}}{{{{\left( {24} \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {20} \right)}^2}}} = 1

For the height of 10 feet, the equation becomes

\begin{gathered} \frac{{{x^2}}}{{{{\left( {24} \right)}^2}}} + \frac{{{{\left( {10} \right)}^2}}}{{{{\left( {20} \right)}^2}}} = 1 \\ \Rightarrow \frac{{{x^2}}}{{576}} + \frac{{100}}{{400}} = 1\,\,\, \Rightarrow \frac{{{x^2}}}{{576}} + \frac{1}{4} = 1 \\ \Rightarrow \frac{{{x^2}}}{{576}} = \frac{3}{4} \\ \Rightarrow x = \pm 12\sqrt 3 \\ \end{gathered}

Thus, the required arch length is 24\sqrt 3 .

Example 2: An asteroid has elliptical orbit with the sun at one focus. Its distance from the sun ranges from 18 million to 182 million miles. Write an equation of the orbit of the asteroid.


app-ellipse-02

Here we have

\begin{gathered} {V_2}{F_2} = 18 = {V_1}{F_1} \\ {V_1}{F_2} = 182 = {V_2}{F_1} \\ \end{gathered}

We also have

\begin{gathered} 2ae = |{F_1}{F_2}| = |{V_1}{F_2} - {V_1}{F_1}| = 182 - 18 = 164 \\ \Rightarrow ae = 82 \\ a = |C{V_2}| = |C{F_2}| - |{F_2}{V_2}| = 82 + 18 = 100 \\ \end{gathered}

Now using the relation to find the value of b, we have

\begin{gathered} {\left( {ae} \right)^2} = {a^2} - {b^2} \\ \Rightarrow {b^2} = {\left( {100} \right)^2} - {\left( {82} \right)^2} = 3276 \\ \end{gathered}

The required equation of the ellipse of the asteroid will be

\begin{gathered} \frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1 \\ \Rightarrow \frac{{{x^2}}}{{10000}} + \frac{{{y^2}}}{{3276}} = 1 \\ \end{gathered}