5.  Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.

     5y^2 - 9x^2 = 36

Answers (1)
P Pankaj Sanodiya

Given a Hyperbola equation,

5y^2 - 9x^2 = 36

Can also be written as 

\frac{5y^2}{36} - \frac{9x^2}{36} = 1

\frac{y^2}{\frac{36}{5}} - \frac{x^2}{4} = 1

\frac{y^2}{(\frac{6}{\sqrt{5}})^2} - \frac{x^2}{2^2} = 1

Comparing this equation with the standard equation of the hyperbola:

\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1

We get,

a=\frac{6}{\sqrt{5}} 

 and b=2

Now, As we know the relation in a hyperbola,

c^2=a^2+b^2

c=\sqrt{a^2+b^2}

c=\sqrt{(\frac{6}{\sqrt{5}})^2+2^2}

c=\sqrt{\frac{56}{5}}

c=2\sqrt{\frac{14}{5}}

Here as we can see from the equation that the axis of the hyperbola is Y-axis. So, 

Coordinates of the foci:

(0,c) \:and\:(0,-c)=\left(0,2\sqrt{\frac{14}{5}}\right)\:and\:\left(0,-2\sqrt{\frac{14}{5}}\right)

The Coordinates of vertices:

(0,a) \:and\:(0,-a)=\left(0,\frac{6}{\sqrt{5}}\right)\:and\:\left(0,-\frac{6}{\sqrt{5}}\right)

The Eccentricity:

e=\frac{c}{a}=\frac{2\sqrt{\frac{14}{5}}}{\frac{6}{\sqrt{5}}}=\frac{\sqrt{14}}{3}

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(4)}{\frac{6}{\sqrt{5}}}=\frac{4\sqrt{5}}{3}

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