Given:
The equation is 5y2 – 9x2 = 36
Let us divide the whole equation by 36, we get
\( \dfrac{5y^2}{36} – \dfrac{9x^2}{36} =\dfrac{ 36}{36}\)
\(\dfrac{ y^2}{(\dfrac{36}{5})} – \dfrac{x^2}{4} = 1\)
On comparing this equation with the standard equation of hyperbola \(\dfrac{ y^2}{a^2} – \dfrac{x^2}{b^2} = 1\)
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 =\(\dfrac{c}{a}\) = (\( \dfrac{2\sqrt{14}}{\sqrt{5}}\))\( \dfrac{6}{\sqrt{5}}\) = \( \dfrac{14}{\sqrt{3}}\)
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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