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  • Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. 9 y ^ 2 - 4 x ^ 2 = 36

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

      9 y^2 - 4 x^2 =36

Answers (1)

best_answer

Given a Hyperbola equation,

9 y^2 - 4 x^2 =36

Can also be written as 

\frac{9y^2}{36} - \frac{4x^2}{36} = 1

\frac{y^2}{2^2} - \frac{x^2}{3^2} = 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=2 and b=3

Now, As we know the relation in a hyperbola,

c^2=a^2+b^2

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

c=\sqrt{2^2+3^2}

c=\sqrt{13}

Hence, 

Coordinates of the foci:

(0,c) \:and\:(0,-c)=(0,\sqrt{13})\:and\:(0,-\sqrt{13})

The Coordinates of vertices:

(0,a) \:and\:(0,-a)=(0,2)\:and\:(0,-2)

The Eccentricity:

e=\frac{c}{a}=\frac{\sqrt{13}}{2}

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(9)}{2}=\frac{18}{2}=9

Posted by

Pankaj Sanodiya

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