Question

Q2.Findthefocus&vertices,eccentricity&latusrectumofhyperbola

5y2
-9x2
=36.​

Answer 1

EXPLANATION.

To find focus , vertices , eccentricity and latus rectum of hyperbola

5y² - 9x² = 36.

General equation of hyperbola = y²/a² - x²/b² = 1.

5y² - 9x² = 36.

y²/36/5 - x²/4 = 1.

(a) vertices = ( 0,a ) and ( 0,-a).

vertices = ( 0,6/√5)  and ( 0,-6/√5).

(b) eccentricity = b² = a² ( e² - 1 ).

⇒ 4 = 36/5 = ( e² - 1 ).

⇒ 20 = 36 ( e² - 1 ).

⇒ 20 = 36e² - 36.

⇒ 20 + 36 = 36e².

⇒ 56 = 36e².

⇒ 14 = 9e².

⇒ e = √14/3.

(c) = Length of latus rectum = 2b²/a = 2(4)/6/√5 = 4√5/3.

(d) = foci = ( 0,c ) and ( 0,-c).

⇒ c² = a² + b².

⇒ c² = 36/5 + 4.

⇒ c² = 56/5.

⇒ c = 2√14/√5.

foci = ( 0,2√14/√5 )  and  ( 0, -2√14/√5).