Ex 10.4, 3 - Chapter 10 Class 11 Conic Sections
Last updated at April 19, 2024 by Teachoo
Hyperbola
Ex 10.4, 4
Ex 10.4, 2
Ex 10.4, 3 Important You are here
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Ex 10.4, 5 Important
Example 14 (i)
Ex 10.4, 7 Important
Example 15
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Ex 10.4, 9 Important
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Ex 10.4, 11 Important
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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, 3 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 9y2 – 4x2 = 36 The given equation is 9y2 – 4x2 = 36 Divide whole equation by 36 9𝑦2 − 4𝑥236 = 3636 9𝑦236 − 436x2 = 1 𝑦24 − 𝑥29 = 1 The above equation of hyperbola is of the form 𝑦2𝑎2 – 𝑥2𝑏2 = 1 ∴ Axis of Hyperbola is y-axis Comparing (1) & (2) a2 = 4 a = 2 & b2 = 9 b = 3 Also, c2 = a2 +b2 Putting value of a2 & b2 c2 = 4 + 9 c2 = 13 c = 𝟏𝟑 Co-ordinates of foci = (0, ±c) = (0, ±13) Foci are (0, 13) & (0, −13) Co-ordinates of vertices = (0, ±a) = (0, ±2) So co-ordinates of vertices are (0, 2) & (0, −2) Eccentricity = e = 𝑐𝑎 = 132 Latus rectum = 2𝑏22 = 2 × 92 = 9