**Ex 11.4, 5**

Last updated at March 9, 2017 by Teachoo

Last updated at March 9, 2017 by Teachoo

Transcript

Ex 11.4, 5 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y2 – 9x2 = 36 Given equation is 5y2 – 9x2 = 36. Dividing whole equation by 36 5𝑦236 − 9𝑥236 = 3636 𝑦2365 − 𝑥24 = 1 The above equation is of the form 𝑦2𝑎2 − 𝑥2𝑏2 = 1 ∴ Axis of hyperbola is y-axis Comparing (1) & (2) a2 = 365 a = 𝟔𝟓 & b2 = 4 b = 2 Also, c2 = a2 + b2 c2 = 365 + 4 c2 = 36 + 205 c2 = 565 c2 = 565 c = 𝟐𝟏𝟒𝟓 Co−ordinate of foci = (0, ± c) = 0, ± 2145 So, co-ordinates of foci are 0, 2145 & 0, −2145 Coordinates of vertices = (0, ±a) = (0, ±65) So, co-ordinates of vertices are 0, 65 & 0,−65 Eccentricity is e = 𝑐𝑎 e = 214565 e = 2145 × 56 = 143 Latus rectum = 2𝑏2𝑎 = 2 × 2265 = 2 × 4 × 56 = 453

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can contact him here.