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

Foci (± 4, 0), the latus rectum is of length 12

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Given, in a hyperbola

Foci (± 4, 0), the latus rectum is of length 12

Here, focii are 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 standard parameter (length of latus rectum and foci) with the given one, we get

$c=4$ and

$\frac{2b^2}{a}=12\Rightarrow 2b^2=12a\Rightarrow b^2=6a$

Now, As we know the relation in a hyperbola

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

$c^2=a^2+6a$

$a^2+6a=4^2$

$a^2+6a=16$

$a^2+8a-2a-16=0$

$(a+8)(a-2)=0$

$a=-8\:or\:2$

Since $a$ can never be negative,

$a=2$

$a^2=4$

$b^2=6a=6(2)=12$

Hence, The Equation of the hyperbola is ;

$\frac{x^2}{4}-\frac{y^2}{12}=1$

Posted by

Pankaj Sanodiya

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