Topic
Write down the canonical equation of the ellipse (E) in each of the following cases:
a) The length of the major axis is 6 and the focal point is \({F_1}\left( { – 2;0} \right)\)
b) Focal length is 12 and eccentricity is \(\frac{3}{5}\)
c) The eccentricity is \(\frac{{\sqrt 5 }}{3}\) and the perimeter of the base rectangle of (E) is 20
Solution method – See details
For ellipse(E): \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) \((0 < b < a)\)
+ Large axis length: \(2a\), minor axis length: \(2b\)
+ Focus \({F_1}( – c;0),{F_2}(c;0)\)
+ Focal length: \(2c = 2\sqrt {{a^2} – {b^2}} \)
+ Miscenter of the ellipse: \(e = \frac{c}{a}\) with \(c = \sqrt {{a^2} – {b^2}} \).
+ The dimensions of the base rectangle are 2a and 2b
Detailed explanation
a) We have: Length of major axis \(2a = 6 \Rightarrow a = 3\) and focus is \({F_1}\left( { – 2;0} \right)\) so \(c = 2 \Rightarrow b = \sqrt {{a^2} – {c^2}} = \sqrt {{3^2} – {2^2}} = \sqrt 5 \)
So ellipse (E): \(\frac{{{x^2}}}{9} + \frac{{{y^2}}}{5} = 1\)
b) We have: Focal length \(2c = 12 \Rightarrow c = 6\)
The eccentricity is equal to \(e = \frac{c}{a} = \frac{6}{a} = \frac{3}{5} \Rightarrow a = 10\)
\( \Rightarrow b = \sqrt {{a^2} – {c^2}} = \sqrt {{{10}^2} – {6^2}} = 8\)
So ellipse (E): \(\frac{{{x^2}}}{{100}} + \frac{{{y^2}}}{{64}} = 1\)
c) False center of \(e = \frac{c}{a} = \frac{{\sqrt 5 }}{3} \Rightarrow \frac{{\sqrt {{a^2} – {b^2} } }}{a} = \frac{{\sqrt 5 }}{3} \Rightarrow \frac{{{a^2} – {b^2}}}{{{a^2}}} = \frac {5}{9}\)
\( \Rightarrow 9{a^2} – 9{b^2} = 5{a^2} \Rightarrow 4{a^2} = 9{b^2} \Rightarrow 2a = 3b\)
With the perimeter of the base rectangle of (E) equal to 20, \(2\left( {2a + 2b} \right) = 20 \Rightarrow a + b = 5\)
\( \Rightarrow \left\{ \begin{array}{l}2a = 3b\\a + b = 5\end{array} \right. \Rightarrow \left\{ \begin{array}{l}a = 3\\b = 2\end{array} \right.\)
So ellipse (E): \(\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1\)
