Q. Suppose X is one end of the major axis of the ellipse $ \frac{x^2}{9} + \frac{y^2}{b^2} = 1 $ (b < 3) and F is that focus which is farther from X. Let Y denote one end of the minor axis. If $FYX$ is a right angled triangle, then the length of the latus rectum of the ellipse is

Solution:

$\Delta FYX$ is a right angle triangle.
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Slope of $ FY\times$ Slope of $XY = - 1$
$\frac{0-b}{-3e -0} \cdot \frac{0-b}{3-0} = -1 \Rightarrow b^{2}= 9e $
We know, $e^{2} = 1 - \frac{b^{2}}{a^{2}} \Rightarrow \frac{b^{4}}{81} = 1-\frac{b^{2}}{9} $
$\Rightarrow b^{4} + 9b^{2} -81 = 0$
$ \Rightarrow b^{2} = \frac{-9\pm \sqrt{81 +\left(4\times1\times81\right)}}{2\times1} $
$\Rightarrow b^{2} = \frac{-9\pm9\sqrt{5}}{2} \Rightarrow b^{2} = \frac{-9 + 9\sqrt{5}}{2} $
Length of latus rectum is $\frac{2b^{2}}{a} $
$= \frac{2\left(-9 +9\sqrt{5}\right)}{2\times3} = 3\left(\sqrt{5}-1\right)$

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