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14.  Find the equations of the hyperbola satisfying the given conditions.

      vertices (± 7,0), e = \frac{4}{3}

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Given, in a hyperbola

vertices (± 7,0), And

 e = \frac{4}{3}

Here, Vertices is  on the X-axis so, the standard equation of the Hyperbola will be ;

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

By comparing the standard parameter (Vertices and eccentricity) with the given one, we get

a=7 and 

e=\frac{c}{a}=\frac{c}{7}=\frac{4}{3}

From here,

c=\frac{28}{3}

Now, As we know the relation  in a hyperbola 

c^2=a^2+b^2

b^2=c^2-a^2

b^2=\left(\frac{28}{3}\right)^2-7^2

b^2=\left(\frac{784}{9}\right)-49

b^2=\left(\frac{784-441}{9}\right)=\frac{343}{9}

Hence, The Equation of the hyperbola is ;

\frac{x^2}{49}-\frac{9y^2}{343}=1

Posted by

Pankaj Sanodiya

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