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Find the coordinate of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.1. \(\frac{x^2}{4} + \frac{y^2}{25} = 1\)2. \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) |
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Answer» 1. Since 25 > 4 the standard equation of the ellipse is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\) ⇒ a2 = 25; b2 = 4 c2 = a2 – b2 = 25 – 4 = 21 ⇒ c = √21 Coordinate of foci are (0, ±√21) Coordinate of vertex are (0, ±5) Length of major axis = 2a = 2 × 5 = 10 Length of minor axis = 2b = 2 × 2 = 4 Eccentricity = e = \(\frac{c}{a}\) =\(\frac{\sqrt{21}}{5}\) Length of latus rectum = \(\frac{2b^2}{a} = \frac{2 \times 4}{5} = \frac{8}{5}.\) 2. Since 16 > 9 the standard equation of the ellipse is \(\frac{x^2}{a} + \frac{y^2}{b^2} = 1\) ⇒ a2 = 16; b2 = 9 c2 = a2 – b2 = 16 – 9 = 7 ⇒ c = √7 Coordinate of foci are (±√7, 0) Coordinate of vertex are (±4, 0) Length of major axis = 2a = 2 × 4 = 8 Length of minor axis = 2b = 2 × 3 = 6 Eccentricity = e = \(\frac{c}{a}\)=\(\frac{\sqrt{7}}{4}\) Length of latus rectum = \(\frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2}.\) |
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