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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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In the parabola $y^2=4ax$ , the length of the chord passing through the vertex and inclined to the x-axis at an angle $\theta$ is

$\begin{array}{1 1}(A)\;4a \cos \theta /\sin ^2 \theta \\(B)\;4a \sin \theta / \cos ^2 \theta \\(C)\;a \sec^2 \theta\\(D)\;a cosec ^2 \theta \end{array}$

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1 Answer

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$y^2=4ax$-----(i)
Vertex $=(0,0)$
$m= \tan \theta$
Hence equation of the chord is $y=(\tan \theta)x$
Put in (i) we have $\tan ^2 \theta (x^2)=4ax$
=> $x= \large\frac{4a}{\tan ^2 \theta}$
$y= \large\frac{4a}{\tan \theta}$
Length of the chord $=\sqrt {\bigg(\large\frac{4a}{\tan 2t} \bigg)^2+\bigg(\large\frac{4a}{\tan \theta}\bigg)^2}$
$\qquad= \large\frac{4a}{\tan ^2 \theta} $$ \sqrt {1+\tan ^2 \theta}$
$\qquad= \large\frac{4a \cos ^2 \theta}{\sin ^2t } \times \large\frac{1}{\cot \theta}$
$\qquad= \large\frac{4a \cos \theta}{\sin ^2 \theta}$
Hence A is the correct answer.
answered Apr 9, 2014 by meena.p
 

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