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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equations of the hyperbola satisfying the given conditions : Foci $(\pm 3\sqrt 5 , 0)$, the latus rectum is of length $8$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{25}+\large\frac{y^2}{20}=1 & \quad (B)\;\large\frac{x^2}{20}+\large\frac{y^2}{25}=1 \\ (C)\;\large\frac{x^2}{25}-\large\frac{y^2}{20}=1 & \quad (D)\;\large\frac{y^2}{25}-\large\frac{x^2}{20}=1 \end {array}$

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1 Answer

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Toolbox:
  • 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 the foci given are $( \pm 3 \sqrt 5 , 0)$ and
The length of the latus rectum is 8
Since the foci is on the x - axis.
$ \therefore $ Equation of the hyperbola is
$ \large\frac{x^2}{a^2}$$-\large\frac{y^2}{b^2}$$=1$
Length of the latus rectum is $ \large\frac{2b^2}{a}$
(i.e) $\large\frac{2 \times b^2}{a}$$=8$
$ \Rightarrow b^2=4a$
Step 2 :
We know that $c^2=a^2+b^2$
(i.e) $(3\sqrt 5 )^2=a^2+4a$
$ \Rightarrow 45 = a^2+4a$
or $a^2+4a-45=0$
On factorizing we get,
$a^2+9a-5a-45=0$
$a(a+9)-5(a+9)=0$
$ \Rightarrow (a-5)(a+9)=0$
$ \therefore a = 5 $ or $ a =-9$
But $a=-9$ is not admissable.
$ \therefore a = 5$
$ \Rightarrow b^2=4a$
$ = 4 \times 5 =20$
Hence the equation of the hyperbola is
$ \large\frac{x^2}{25}$$-\large\frac{y^2}{20}=1$
answered Jul 21, 2014 by thanvigandhi_1
 

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