The equation is 5y2 – 9x2 = 36
Let us divide the whole equation by 36, we get
\( \dfrac{5y^2}{9}- \dfrac{9x^2}{36}=1\)
= \( \dfrac{36}{36}\)
\( \dfrac{y^2}{\dfrac{36}{5}}-\dfrac{x^2}{4}=1\)
On comparing this equation with the standard equation of hyperbola
\( \dfrac{x^2}{a^2}\) +\( \dfrac{y^2}{b^2}\) = 1,
We get a = \( \dfrac{6}{\sqrt{5}}\) and b = 2,
It is known that, a2 + b2 = c2
So,
c2 = \( \dfrac{36}{5}\) + 4
c2 = \( \dfrac{56}{5}\)
c = \( \sqrt{\dfrac{56}{5}}\)
= \( \dfrac{2\sqrt{14}}{\sqrt{5}}\)
Then,
The coordinates of the foci are
(0, \( \dfrac{2\sqrt{14}}{\sqrt{5}}\)) and (0, – \( \dfrac{2\sqrt{14}}{\sqrt{5}}\)).
The coordinates of the vertices are
(0,\( \dfrac{6}{\sqrt{5}}\)) and (0, \( - \dfrac{6}{\sqrt{5}}\)).
Eccentricity, e =\( \dfrac{c}{a}\)
= \( \dfrac{ \dfrac{2\sqrt{14}}{\sqrt{5}}}{\dfrac{6}{\sqrt{5}}}\)
= \(\dfrac{\sqrt{14}}{3}\)
Length of latus rectum = \( \dfrac{2b^2}{a}\)
= \( \dfrac{(2×2^2)}{\dfrac{6}{\sqrt{5}}}\)
= \( \dfrac{(2×4)}{\dfrac{6}{\sqrt{5}}}\)
= \(\dfrac{4\sqrt{5}}{3}\)
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