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Question Number 9163 by tawakalitu last updated on 21/Nov/16

show that the ellipse with e = \frac{\sqrt{5}}{3},

focus (0, 2) and directrix x = - \frac{4 \sqrt{5}}{3}

has the equation : \frac{{(x - \sqrt{5})}^{2}}{9} + \frac{{(y - 2)}^{2}}{4} = 1

Commented by sandy_suhendra last updated on 22/Nov/16

equation of an ellipse is \frac{{(x - h)}^{2}}{a^{2}} + \frac{{(y - k)}^{2}}{b^{2}} = 1

where

a = major axis

b = minor axis

(h, k) is the centre of ellipse

linear eccentricity = c = \sqrt{a^{2} - b^{2}}

eccentricity = e = \frac{c}{a}

directrix \Rightarrow x = h \pm \frac{a}{e}

focal points : (h \pm c, k)

e = \frac{\sqrt{5}}{3} \Rightarrow \frac{\sqrt{5}}{3} = \frac{c}{a} \Rightarrow c = \frac{a \sqrt{5}}{3}

focus : (0, 2) \Rightarrow h - c = 0 \Rightarrow h = c = \frac{a \sqrt{5}}{3} (h + c is imposible = 0)

k = 2

directrix \Rightarrow x = - \frac{4 \sqrt{5}}{5} (I think that is 5 not 3)

- \frac{4 \sqrt{5}}{5} = h - \frac{a}{e}

- \frac{4 \sqrt{5}}{5} = \frac{a \sqrt{5}}{3} - \frac{a}{\sqrt{5} / 3}

- \frac{4 \sqrt{5}}{5} = \frac{a \sqrt{5}}{3} - \frac{3 a \sqrt{5}}{5} (multiplied by \frac{15}{\sqrt{5}})

- 12 = 5 a - 9 a

a = 3 \Rightarrow c = h = \sqrt{5}

c^{2} = a^{2} - b^{2}

5 = 9 - b^{2}

b = 2

substitute : a = 3, b = 2, h = \sqrt{5}, k = 2

to the equation of ellipse, so we have :

\frac{{(x - \sqrt{5})}^{2}}{9} + \frac{{(y - 2)}^{2}}{4} = 1

Commented by tawakalitu last updated on 22/Nov/16

Wow, this is great . i really appreciate sir .

God bless you .