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