Let H be the hyperbola, whose foci are 1 ± 2 , 0 and eccentricity is 2 . Then the length of its latus rectum is:

Question

TestHub · Accepted Answer

Given the foci points of a hyperbola are $(1 \pm \sqrt{2}, 0)$

So, $S (1 + \sqrt{2}, 0), S^{'} (1 - \sqrt{2}, 0)$

$S S^{'} = 2 ae = (1 + \sqrt{2}) - (1 - \sqrt{2})$

$\Rightarrow 2 a e = 2 \sqrt{2}$

$\Rightarrow 2 \times a \times \sqrt{2} = 2 \sqrt{2}, (given e = \sqrt{2})$

$\Rightarrow a = 1$

Since $b^{2} = a^{2} (e^{2} - 1)$

$\Rightarrow b^{2} = 1 (2 - 1) = 1$

So, the latus rectum of a hyperbola is given by,

L.R. $= \frac{2 b^{2}}{a}$

L.R. $= 2$

Mathematics - Hyperbola Question with Solution | TestHub