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

$5y^2 - 9x^2 = 36$

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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 :

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

Posted by

Pankaj Sanodiya

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