Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:
(i) 4x² + 9y² = 1
(ii) 5x² + 4y² = 1
(iii) 4x² + 3y² = 1
(iv) 25x² + 16y² = 1600.
(v) 9x² + 25y² = 225

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Solution

$(i) 4 x^{2} + 9 y^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{4}} + \frac{y^{2}}{\frac{1}{9}} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = \frac{1}{4} and b^{2} = \frac{1}{9}, i . e . a = \frac{1}{2} and b = \frac{1}{3} . Clearly a > b Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{1}{9}}{\frac{1}{4}}} \Rightarrow e = \sqrt{1 - \frac{4}{9}} \Rightarrow e = \frac{\sqrt{5}}{3} Coordinates of the foci = (\pm a e, 0) = (\pm \frac{\sqrt{5}}{6}, o) Length of the latus rectum= \frac{2 b^{2}}{a} = \frac{2 \times \frac{1}{9}}{\frac{1}{2}} = \frac{4}{9}$
$(ii) 5 x^{2} + 4 y^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{5}} + \frac{y^{2}}{\frac{1}{4}} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = \frac{1}{5} and b^{2} = \frac{1}{4}, i . e . a = \frac{1}{\sqrt{5}} and b = \frac{1}{2} . Clearly b > a Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{1}{5}}{\frac{1}{4}}} \Rightarrow e = \sqrt{1 - \frac{4}{5}} \Rightarrow e = \frac{1}{\sqrt{5}} Coordinates of the foci = (0, \pm b e) = (0, \pm \frac{1}{2 \sqrt{5}}) Length of the latus rectum= \frac{2 a^{2}}{b} = \frac{2 \times \frac{1}{5}}{\frac{1}{2}} = \frac{4}{5}$

$(iii) 4 x^{2} + 3 y^{2} = 1 \Rightarrow \frac{x^{2}}{\frac{1}{4}} + \frac{y^{2}}{\frac{1}{3}} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = \frac{1}{4} and b^{2} = \frac{1}{3}, i . e . a = \frac{1}{2} and b = \frac{1}{\sqrt{3}} . Clearly, b > a Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{\frac{1}{4}}{\frac{1}{3}}} \Rightarrow e = \sqrt{1 - \frac{3}{4}} \Rightarrow e = \frac{1}{2} Coordinates of the foci = (0, \pm b e) = (0, \pm \frac{1}{2 \sqrt{3}}) Length of the latus rectum= \frac{2 a^{2}}{b} = \frac{2 \times \frac{1}{4}}{\frac{1}{\sqrt{3}}} = \frac{\sqrt{3}}{2}$

$(iv) 25 x^{2} + 16 y^{2} = 1600 \Rightarrow \frac{x^{2}}{64} + \frac{y^{2}}{100} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = 64 and b^{2} = 100, i . e . a = 8 and b = 10 . Clearly, b > a Now, e = \sqrt{1 - \frac{a^{2}}{b^{2}}} \Rightarrow e = \sqrt{1 - \frac{64}{100}} \Rightarrow e = \sqrt{\frac{36}{100}} \Rightarrow e = \frac{6}{10} or \frac{3}{5} Coordinates of the foci = (0, \pm 6) Length of latus rectum= \frac{2 a^{2}}{b} = \frac{2 \times 64}{10} = \frac{64}{5}$
$(v) 9 x^{2} + 25 y^{2} = 225 \Rightarrow \frac{x^{2}}{25} + \frac{y^{2}}{9} = 1 This is of the form \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1, where a^{2} = 25 and b^{2} = 9, i . e . a = 5 and b = 3 . Clearly, a > b Now, e = \sqrt{1 - \frac{b^{2}}{a^{2}}} \Rightarrow e = \sqrt{1 - \frac{9}{25}} \Rightarrow e = \sqrt{\frac{16}{25}} \Rightarrow e = \frac{4}{5} Coordinates of the foci = (\pm a e, 0) = (\pm 4, 0) Length of latus rectum= \frac{2 b^{2}}{a} = \frac{2 \times 9}{5} = \frac{18}{5}$

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Find the eccentricity, coordinates of foci, length of the latus-rectum of the following ellipse:
(i) $4 x^{2} + 9 y^{2} = 1$
(ii) $5 x^{2} + 4 y^{2} = 1$
(iii) $4 x^{2} + 3 y^{2} = 1$
(iv) $25 x^{2} + 16 y^{2} = 1600$
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