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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The length of a focal chord of the parabola $y^2=4ax$ making an angle $\theta$ with the axis of the parabola is

$\begin{array}{1 1}(A)\;4a \;cosec ^2 \theta \\(B)\;4a\; \sec^2 \theta \\(C)\;a \;cosec^2 \theta \\(D)\;none\; of\; these \end{array}$

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1 Answer

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Length of the focal chord if
$a \sqrt {(t^2 -\large\frac{1}{t^2})^2 +4 (t+\frac{1}{t})^2}$
$\qquad= a \bigg( t+\large\frac{1}{t} \bigg)^2$
Now $\tan \theta =\large\frac{2( t +\large\frac{1}{t})}{t^2-\frac{1}{t^2}}$
$\qquad= \large\frac{2}{t-\large\frac{1}{t^2}}$
=> $t- \large\frac{1}{t}$$= 2 \cot \theta$
=> $ \bigg(t +\large\frac{1}{t}\bigg)^2 =\bigg(t -\large\frac{1}{t} \bigg)^2 +4$
$\qquad= 4 \cot ^2 \theta+4$
$\qquad= 4 a \;cosec ^2 \theta$
Hence A is the correct answer.
answered Apr 9, 2014 by meena.p
 

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