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 we have length of major axis is given as 2a = 48\,\, \Rightarrow a = 24 and height of a semi-ellipse is given as b = 20.
Now standard equation of 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 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 elliptic 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}


Also we 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}


Required equation of ellipse of 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}

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