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# Find the equations of the hyperbola satisfying the given conditions : Foci $(\pm 4, 0)$, the latus rectum is of length $12$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{4}+\large\frac{y^2}{12}=1 & \quad (B)\;\large\frac{x^2}{12}+\large\frac{y^2}{4}=1 \\ (C)\;\large\frac{x^2}{4}-\large\frac{y^2}{12}=1 & \quad (D)\;\large\frac{y^2}{12}-\large\frac{x^2}{4}=1 \end {array}$

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## 1 Answer

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• Standard equation of a hyperbola where transverse axis is along x - axis is $\large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2}$$=1$
• Standard equation of a hyperbola whose transverse axis is along y - axis is $\large\frac{y^2}{a^2} $$- \large\frac{x^2}{b^2}$$=1$
• $c^2=a^2+b^2$, where $c$ is the foci
• Eccentricity $e = \large\frac{c}{a}$
• Length of the latus rectum $= \large\frac{2b^2}{a}$
Step 1 :
The coordinates of foci are $( \pm 4, 0)$
The length of the latus rectum is 12.
Since the foci are on the x - axis, the equation of the hyperbola is of the form $\large\frac{x^2}{a^2} - \large\frac{y^2}{b^2}=1$
Length of the latus rectum is $\large\frac{2b^2}{a}$
Since $2b^2=12 \Rightarrow b^2 = 6a$ and $c = 4$
Step 2 :
We know $c^2=a^2+b^2$
Substituting for c and a we get,
$16=a^2+16a^2$
$\Rightarrow a^2+6a-16=0$
On factorizing we get,
$a^2+8a-2a-16=0$
$(a+8)(a-2)=0$
$a=-8,2$
$a=-8$ is not admissable.
Hence a = 2
$\therefore b^2=6a$
$= 6 \times 2 =12$
Thus the equation of the hyperbola is
$\large\frac{x^2}{4}-\large\frac{y^2}{12}=1$
answered Jul 21, 2014

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