## Solving Lesson 1 Page 48 Math Learning Topic 10 – Kite>

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): $$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$$