Let a and b respectively be the semitransvers | vedclass

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Question

$S\left( {5,0} \right)\,$ is focus $ \Rightarrow ae = 5$ (focus)          ......$(1)$

$x = \frac{a}{5} \Rightarrow \f

Vedclass · Accepted Answer

$-7$

Vedclass · Answer

$S\left( {5,0} \right)\,$ is focus $ \Rightarrow ae = 5$ (focus)          ......$(1)$

$x = \frac{a}{5} \Rightarrow \frac{a}{e} = \frac{9}{5}$  (directrix)        .....$(2)$        

$\left( 1 \right) \left( 2 \right) \Rightarrow {a^2} = 9$

$\left( 1 \right) \Rightarrow \left( e \right) = \frac{5}{3}$

${b^2} = {a^2}\left( {{e^2} - 1} \right) \Rightarrow {b^2} = 16$

${a^2} + {b^2} = 9 - 16 =  - 7$

Let $a$ and $b$ respectively be the semitransverse and semi-conjugate axes of a hyperbola whose eccentricity satisfies the equation $9e^2 - 18e + 5 = 0.$ If $S(5, 0)$ is a focus and $5x = 9$ is the corresponding directrix of this hyperbola, then $a^2 - b^2$ is equal to

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