Let the foci of the ellipse x216+y27=1 and the hyperbola x2144−y2α=136 coincide. Then the length of the latus rectum of the hyperbola is _____

x216+y27=1⇒e=34, foci=(±3,0)Eccentricity of hyperbola is 144+α12Foci (2144+α12,0) ∴144+α6=3 144+α=324 α=324−144=180 L.L.R=2b2a=2(180536)2=5

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Let the foci of the ellipse $\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1$ and the hyperbola $\frac{x^{2}}{144} - \frac{y^{2}}{α} = \frac{1}{36}$ coincide. Then the length of the latus rectum of the hyperbola is _____

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answer is 5.

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Detailed Solution

$\frac{x^{2}}{16} + \frac{y^{2}}{7} = 1 \Rightarrow e = \frac{3}{4}, f o c i = (\pm 3, 0)$
Eccentricity of hyperbola is $\frac{\sqrt{144 + α}}{12}$
Foci $(\frac{2 \sqrt{144 + α}}{12}, 0)$
$∴ \frac{\sqrt{144 + α}}{6} = 3$
$144 + α = 324$
$α = 324 - 144 = 180$
$L . L . R = \frac{2 b^{2}}{a} = \frac{2 (\frac{180^{5}}{36})}{2} = 5$

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