## Solve Lesson 2 Page 48 Math Study Topic 10 – Kite>

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 $$\frac{{\sqrt {10} }}{5}$$