Let an ellipse (equation not preserved) have eccentricity 1√2 and latus rectum √14.

Question

Let an ellipse (equation not preserved) have eccentricity $\dfrac{1}{\sqrt2}$ and latus rectum $\sqrt{14}$. Then the square of the eccentricity of the associated hyperbola is:

Accepted Answer

$e^2=\dfrac12=1-\dfrac{b^2}{a^2}\Rightarrow b^2=\dfrac{a^2}{2}$. Latus rectum: $\dfrac{2b^2}{a}=\sqrt{14}\Rightarrow a\cdot\dfrac1{}=\sqrt{14}\Rightarrow a=\sqrt{14},\ b^2=7$.

For the hyperbola $\dfrac{x^2}{14}-\dfrac{y^2}{7}=1$: $e_H^2=1+\dfrac{7}{14}=\dfrac32$.

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