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Home  >>  CBSE XI  >>  Math  >>  Conic Sections
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Find the equations of the hyperbola satisfying the given conditions : vertices $(\pm 7,0 ) , e = \large\frac{4}{3}$

$\begin {array} {1 1} (A)\;\large\frac{x^2}{49}-\large\frac{y^2}{343}=1 & \quad (B)\;\large\frac{9x^2}{343}-\large\frac{y^2}{49}=1 \\ (C)\;\large\frac{x^2}{49}+\large\frac{9y^2}{343}=1 & \quad (D)\;\large\frac{9x^2}{343}+\large\frac{y^2}{49}=1 \end {array}$

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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 given vertices is $ ( \pm 7, 0) $ and eccentricity $e = \large\frac{4}{3}$
The vertices lie on the x - axis.
Hence the equation of the hyperbola is of the form $ \large\frac{x^2}{a^2} - \large\frac{y^2}{b^2} = 1$
Since the vertices are $( \pm 7, 0)$
$ a = 7 $
$ e = \large\frac{4}{3}$ ( Hint : $e = \large\frac{c}{a} $)
Step 2 :
(i.e) $ \therefore \large\frac{c}{a} = \large\frac{4}{3}$
$ \Rightarrow \large\frac{c}{7} = \large\frac{4}{3}$
$ \Rightarrow c = \large\frac{28}{3}$
Step 3 :
$c^2=a^2+b^2$
$\therefore 7^2+b^2= \bigg( \large\frac{28}{3} \bigg)^2$
$ \therefore b^2 = \large\frac{784}{9} $$- 49 = \large\frac{784-441}{9}$
$ b^2 = \large\frac{343}{9}$
Hence the required equation of the parabola is $ \large\frac{x^2}{49} - \large\frac{y^2}{\Large\frac{343}{9}}=1$
(i.e) $ \large\frac{x^2}{49} - \large\frac{9y^2}{343}$$=1$
answered Jul 21, 2014 by thanvigandhi_1
edited Jul 21, 2014 by thanvigandhi_1
 

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