Question

Question 5 Find the coordinates of the foci and the vertices, the eccentricity, and the length of the latus rectum of the hyperbola 5y^2 – 9x^2 = 36

Class X1 - Maths -Conic Sections Page 262

Answer 1

5y² - 9x² = 36
divided by 36 from both sides,
5y²/36 - 9x²/36 = 36/36
y²/{36/5} - x²/{36/9} = 1
y²/{6/√5}² - x²/(2)² = 1 --------------(1)

concept :- if equation of hyperbola will be of the form y²/a² - x²/b² = 1
then,
i) transverse axis along Y-axis.
ii) vertices ( 0, ± a)
iii) foci ( 0, ± c) where c² = a² + b²
iv) eccentricity (e) = c/a
v) Latusrectum = 2b²/a

see equation (1) , you observe that this equation similar as y²/a² - x²/b² = 1
where , a = 6/√5 and b = 2
then, c² = a² + b² = 36/5 + 4 = 56/5
c = 2√14/√5

hence, foci ( 0, ± c) = ( 0, ±2√14/√5)
vertices ( 0, ± a) = ( 0, ± 6/√5)
eccentricity ( e ) = c/a = {2√14/√5}/{6/√5} =√14/3
Latusrectum = 2b²/a = 2 × 4/{6/√5} = 4√5/3