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# Find the equations of the hyperbola satisfying the given conditions : vertices $(\pm 7,0 ) , e = \large\frac{4}{3}$

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

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$
edited Jul 21, 2014