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

      49y^2 - 16x^2 = 784

Answers (1)
P Pankaj Sanodiya

Given a Hyperbola equation,

49y^2 - 16x^2 = 784

Can also be written as

 \frac{49y^2}{784} - \frac{16x^2}{784} = 1

\frac{y^2}{16} - \frac{x^2}{49} = 1

\frac{y^2}{4^2} - \frac{x^2}{7^2} = 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=4 and b=7

Now, As we know the relation in a hyperbola,

c^2=a^2+b^2

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

c=\sqrt{4^2+7^2}

c=\sqrt{65}

Therefore,

Coordinates of the foci:

(0,c) \:and\:(0,-c)=(0,\sqrt{65})\:and\:(0,-\sqrt{65})

The Coordinates of vertices:

(0,a) \:and\:(0,-a)=(0,4)\:and\:(0,-4)

The Eccentricity:

e=\frac{c}{a}=\frac{\sqrt{65}}{4}

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(49)}{4}=\frac{98}{4}=\frac{49}{2}

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