Last updated at Feb. 6, 2020 by Teachoo
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Example 14 Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the hyperbolas: (i) x2/9 โ y2/16 = 1, The given equation is ๐ฅ2/9 โ ๐ฆ2/16 = 1 The above equation is of the form ๐ฅ2/๐2 โ ๐ฆ2/๐2 = 1 Comparing (1) & (2) a2 = 9 a = 3 & b2 = 16 b = 4 Also, c2 = a2 + b2 c2 = 9 + 16 c2 = 25 c = 5 So, Co-ordinate of foci = (ยฑc, 0) = (ยฑ5, 0) Thus, Co-ordinate of foci are (5, 0) & (โ5, 0) Vertices = (ยฑa, 0) = (ยฑ3, 0) Thus, Vertices are (3, 0) & (โ3, 0) Eccentricity e = ๐/๐ = ๐/๐ Latus rectum = 2๐2/๐ = (2 ร 16)/3 = ๐๐/๐ Example 14 Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the hyperbolas: (ii) y2 โ 16x2 = 16 The given equation is y2 โ 16x2 = 16 Divide whole equation by 16 (๐ฆ2โ16๐ฅ2)/16 = 16/16 ๐ฆ2/16 โ ๐ฅ2/1 = 1 The above equation is of the form ๐ฆ^2/๐^2 โ ๐ฅ^2/๐^2 = 1 Comparing (1) & (2) a2 = 16 a = 4 & b2 = 1 b = 1 Also , c2 = a2 + b2 c2 = 16 + 1 c2 = 17 c = โ๐๐ Co-ordinate of foci = (0, ยฑc) = (0, ยฑโ๐๐) So, co-ordinates of foci are (0, โ17) & (0, โโ17) Vertices = (0, ยฑa) = (0, ยฑ4) So, vertices are (0, 4) & (0, โ4) Eccentricity e = ๐/๐ = โ๐๐/๐ Latus rectum = 2๐2/๐ = (2 ร 1)/4 = ๐/๐
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