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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the length of lactus rectum and end points of lactus rectum for a ellipse $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$?

$\begin{array}{1 1}(a)\;\large\frac{2b^2}{a}\\(b)\;\large\frac{b^2}{a}\\(c)\;\large\frac{4b^2}{a}\\(d)\;\large\frac{2a^2}{b}\end{array}$

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1 Answer

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Latus rectum is a perpendicular line passing through focus hence let the end points be $L(ae,k)$ & $L'(ae,-k)$.
Length of latus rectum LL'=$2k$
$\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
$\large\frac{a^2e^2}{a^2}+\frac{k^2}{b^2}$$=1$
$k^2=b^2(1-e^2)$
$\;\;\;\;=b^2(\large\frac{b^2}{a^2})$
$b^2=a^2(1-e^2)$
$k=\large\frac{b^2}{a}$
$2k=\large\frac{2b^2}{a}=$$LL'$(length of lactus rectum)
Hence (a) is the correct answer.
answered Feb 7, 2014 by sreemathi.v
edited Sep 28, 2014 by sharmaaparna1
 

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