Hyperbola in conic sections

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$If the distances of one focus of hyperbola from its directrices are 5 and 3, then eccentricity of hyperbola is$

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Solution

$According to the question,$
$\begin{aligned} a e - \frac{a}{e} = 3 - - - - (1) \\ a e + \frac{a}{e} = 5 - - - - (2) \end{aligned}$
$\begin{aligned} Solving (1) and (2), we get \\ \begin{array}{l} a e = 4 and \frac{a}{e} = 1 \\ So, a^{2} = 4 \Rightarrow a = 2 \Rightarrow e = 2 \end{array} \end{aligned}$

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Hyperbola in conic sections

Remember concepts with our Masterclasses.

If the distances of one focus of hyperbola from its directrices are 5 and 3, then eccentricity of hyperbola is

According to the question,ae−ae=3----(1)ae+ae=5----(2) Solving (1) and (2), we get ae=4 and ae=1 So, a2=4⇒a=2⇒e=2

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$If the distances of one focus of hyperbola from its directrices are 5 and 3, then eccentricity of hyperbola is$