Let the ellipse E: frac{x^{2}}{144}+frac{y^{2 | vedclass

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(A) The equation of the ellipse is $\frac{x^{2}}{144}+\frac{y^{2}}{169}=1$. Here $a^{2}=144$ and $b^{2}=169$. Since $b^{2} > a^{2}$, the foci lie on t

Vedclass · Accepted Answer

$296$

Vedclass · Answer

(A) The equation of the ellipse is $\frac{x^{2}}{144}+\frac{y^{2}}{169}=1$. Here $a^{2}=144$ and $b^{2}=169$. Since $b^{2} > a^{2}$, the foci lie on the $y$-axis.For the ellipse, $e_{E} = \sqrt{1 - \frac{a^{2}}{b^{2}}} = \sqrt{1 - \frac{144}{169}} = \sqrt{\frac{25}{169}} = \frac{5}{13}$.The foci are $(0, \pm be_{E}) = (0, \pm 13 	imes \frac{5}{13}) = (0, \pm 5)$.The hyperbola is $\frac{y^{2}}{\lambda^{2}} - \frac{x^{2}}{16} = 1$. Here $a^{2} = \lambda^{2}$ and $b^{2} = 16$.The foci are $(0, \pm ae_{H})$, where $ae_{H} = \sqrt{a^{2} + b^{2}} = \sqrt{\lambda^{2} + 16}$.Equating the foci: $\sqrt{\lambda^{2} + 16} = 5$ $\Rightarrow \lambda^{2} + 16 = 25$ $\Rightarrow \lambda^{2} = 9$ $\Rightarrow \lambda = 3$.The eccentricity $e = \frac{\sqrt{a^{2} + b^{2}}}{a} = \frac{5}{3}$.The length of the latus rectum $L = \frac{2b^{2}}{a} = \frac{2(16)}{3} = \frac{32}{3}$.Thus, $24(e+L) = 24(\frac{5}{3} + \frac{32}{3}) = 24(\frac{37}{3}) = 8 	imes 37 = 296$.

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:

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