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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. x ^ 2 / 16 - y ^ 2 / 9 = 1

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

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

Answers (1)

best_answer

Given a Hyperbola equation,

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

Can also be written as 

\frac{x^2}{4^2} - \frac{y^2}{3^2} = 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=4 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{4^2+3^2}

c=5

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

Coordinates of the foci:

(c,0) \:and\:(-c,0)=(5,0)\:and\:(-5,0)

The Coordinates of vertices:

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

The Eccentricity:

e=\frac{c}{a}=\frac{5}{4}

The Length of the latus rectum :

\frac{2b^2}{a}=\frac{2(3)^2}{4}=\frac{18}{4}=\frac{9}{2}

Posted by

Pankaj Sanodiya

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