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

$16x^2 - 9y^2 = 576$

Answers (1)

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Given a Hyperbola equation,

$16x^2 - 9y^2 = 576$

Can also be written as

$\frac{16x^2}{576} - \frac{9y^2}{576} = 1$

$\frac{x^2}{36} - \frac{y^2}{64} = 1$

$\frac{x^2}{6^2} - \frac{y^2}{8^2} = 1$

Comparing this equation with the standard equation of the hyperbola:

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

We get,

$a=6$ and $b=8$

Now, As we know the relation in a hyperbola,

$c^2=a^2+b^2$

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

$c=\sqrt{6^2+8^2}$

$c=10$

Therefore,

Coordinates of the foci:

$(c,0) \:and\:(-c,0)=(10,0)\:and\:(-10,0)$

The Coordinates of vertices:

$(a,0) \:and\:(-a,0)=(6,0)\:and\:(-6,0)$

The Eccentricity:

$e=\frac{c}{a}=\frac{10}{6}=\frac{5}{3}$

The Length of the latus rectum :

$\frac{2b^2}{a}=\frac{2(8)^2}{6}=\frac{128}{6}=\frac{64}{3}$

Posted by

Pankaj Sanodiya

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