Let the ellipse $E:\frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$ and the hyperbola $H:\frac{x^{2}}{16}

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Let the ellipse $E:\frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$ and the hyperbola $H:\frac{x^{2}}{16} - \frac{y^{2}}{\lambda^{2}} = - 1$ have the same foci. If $e$ and $L$ respectively denote the eccentricity and the length of the latus rectum of H , then the value of $24(e + L)$ is :

Accepted Answer

Equation of hyperbola : $\frac{y^{2}}{\lambda^{2}} - \frac{x^{2}}{16} = 1$

Equation of ellipse : $\frac{x^{2}}{144} + \frac{y^{2}}{169} = 1$

$$e' = \sqrt{1 - \frac{144}{169}} = \frac{5}{13} $$focus $\Rightarrow (0,5)$

$${\Rightarrow \lambda\sqrt{1 + \frac{16}{\lambda^{2}}} = 5 }{\Rightarrow \lambda^{2} + 16 = 25 }{\lambda = 3 }$$Eccentricity of hyperbola $= \sqrt{1 + \frac{16}{\lambda^{2}}} = \frac{5}{3}$

Length of latus rectum of hyperbola $= \frac{2(16)}{3} = \frac{32}{3}$

$$24(e + \mathcal{l}) = 24\left\lbrack \frac{5}{3} + \frac{32}{3} \right\rbrack = 8 \times 37 = 296$$

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