The eccentricity e of a hyperbola can never b | vedclass

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Question

(B) For any hyperbola,the eccentricity $e$ must satisfy the condition $e > 1$.We evaluate the given options:$A) \sqrt{\frac{9}{5}} \approx 1.34 > 1$$B

Vedclass · Accepted Answer

$2\sqrt{\frac{1}{9}}$

Vedclass · Answer

(B) For any hyperbola,the eccentricity $e$ must satisfy the condition $e > 1$.We evaluate the given options:$A) \sqrt{\frac{9}{5}} \approx 1.34 > 1$$B) 2\sqrt{\frac{1}{9}} = 2 	imes \frac{1}{3} = \frac{2}{3} \approx 0.67 $C) 3\sqrt{\frac{1}{8}} = 3 	imes \frac{1}{2\sqrt{2}} = \frac{3}{2.828} \approx 1.06 > 1$$D) 2 > 1$Since the eccentricity of a hyperbola cannot be less than or equal to $1$,the value $\frac{2}{3}$ is impossible. Thus,option $B$ is the correct answer.

The eccentricity $e$ of a hyperbola can never be equal to which of the following values?

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