**Topic**

Find the eccentricity of the ellipse (E) in each of the following cases:

a) The length of the major semi-axis is twice the length of the minor semi-axis

b) The distance from a vertex on the major axis to a vertex on the minor axis is equal to the focal length

**Solution method – See details**

For ellipse(E): \(\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\) \((0 < b < a)\)

+ Length of semi-major axis: \(a\), length of semi-major axis: \(b\)

+ Focal length \(2c\) with \(c = \sqrt {{a^2} – {b^2}} \)

+ 4 vertices are \({A_1}\left( { – a;0} \right),{A_2}\left( {a;0} \right),{B_1}\left( {0; – b} \ right),{B_2}\left( {0;b} \right).\)

+ Miscenter \(e = \frac{c}{a}\)

**Detailed explanation**

a) We have: The length of the major semi-axis is twice the length of the minor semi-axis, or \(a = 2b \Rightarrow c = \sqrt {{a^2} – {b^2}} = \sqrt {{a^) 2} – {{\left( {\frac{a}{2}} \right)}^2}} = \sqrt {\frac{3}{4}{a^2}} = \frac{{a \sqrt 3 }}{2}\)

The eccentricity of the ellipse is \(e = \frac{c}{a} = \frac{{\frac{{a\sqrt 3 }}{2}}}{a} = \frac{{\sqrt 3 }} {2}\)

b) Suppose the ellipse has a vertex on the major axis \(A\left( {a;0} \right)\left( {a > 0} \right)\) and a vertex on the minor axis \(B\) left( {0;b} \right)\left( {b > 0} \right)\)

According to the problem we have: \(AB = 2c = 2\sqrt {{a^2} – {b^2}} \)

\(\begin{array}{l} \Rightarrow \sqrt {{{\left( {0 – a} \right)}^2} – {{\left( {b – 0} \right)}^2} } = 2\sqrt {{a^2} – {b^2}} \Rightarrow {a^2} + {b^2} = 4\left( {{a^2} – {b^2}} \ right)\\ \Rightarrow 3{a^2} = 5{b^2} \Rightarrow {b^2} = \frac{3}{5}{a^2} \Rightarrow {c^2} = {a ^2} – {b^2} = {a^2} – \frac{3}{5}{a^2} = \frac{2}{5}{a^2}\\ \Rightarrow \frac{ {{c^2}}}{{{a^2}}} = \frac{2}{5} \Rightarrow e = \frac{c}{a} = \sqrt {\frac{2}{5} } = \frac{{\sqrt {10} }}{5}\end{array}\)

So the ellipse has a false center of \(\frac{{\sqrt {10} }}{5}\)