If the eccentricity of a hyperbola frac{{{x^2 | vedclass

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Question

given hyperbola is 

$\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{{b^2}}} = 1$

Since this passes through $(K,2)$, therfore

$\frac{{{K^2}}}{9} - \

Vedclass · Accepted Answer

$18$

Vedclass · Answer

given hyperbola is 

$\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{{b^2}}} = 1$

Since this passes through $(K,2)$, therfore

$\frac{{{K^2}}}{9} - \frac{4}{{{b^2}}} = 1\,\,\,\,\,\,....\left( 1 \right)$

Also, given  $e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}  = \frac{{\sqrt {13} }}{3}$

$ \Rightarrow \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}  = \frac{{\sqrt {13} }}{3} \Rightarrow 9 + {b^2} = 13$

$ \Rightarrow b =  \pm 2$

Now, from $e{q^n}$ $(1)$, we have 

$\frac{{{K^2}}}{9} - \frac{4}{4} = 1$        ($\because$  $b =  \pm 2$)

$ \Rightarrow {K^2} = 18$

If the eccentricity of a hyperbola $\frac{{{x^2}}}{9} - \frac{{{y^2}}}{{{b^2}}} = 1,$ which passes through $(K, 2),$ is $\frac{{\sqrt {13} }}{3},$ then the value of $K^2$ is

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