Ex 10.4, 2 - Chapter 10 Class 11 Conic Sections
Last updated at April 19, 2024 by Teachoo
Hyperbola
Ex 10.4, 4
Ex 10.4, 2 You are here
Ex 10.4, 3 Important
Ex 10.4, 6
Ex 10.4, 5 Important
Example 14 (i)
Ex 10.4, 7 Important
Example 15
Ex 10.4, 8
Ex 10.4, 9 Important
Ex 10.4, 10
Ex 10.4, 11 Important
Ex 10.4, 12 Important
Ex 10.4, 13
Example 16 Important
Ex 10.4, 14 Important
Ex 10.4, 15 Important
Last updated at April 19, 2024 by Teachoo
Ex 10.4, 2 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola y29 - x227 = 1 Given equation is 𝑦29 – 𝑥227 = 1 The above equation of hyperbola is of the form 𝑦2𝑎2 – 𝑥2𝑏2 = 1 ∴ Axis of Hyperbola is y-axis Comparing (1) & (2) a2 = 9 a = 3 & b2 = 27 b = 3 × 3 × 3 b = 3𝟑 Also, c2 = a2 +b2 Putting value of a2 & b2 c2 = 9 + 27 c2 = 36 c = 6 Co-ordinates of foci = (0, ±c) = (0, ±6) So, co-ordinate of foci are (0, 6) & (0, −6) Co-ordinates of Vertices = (0, ±a) = (0, ±3) So, co-ordinates of vertices are (0, 3) & (0, −3) Eccentricity = e = 𝑐𝑎 = 63 = 2 Latus rectum = 2𝑏2𝑎 = 2 × 273 = 18