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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The angle between the tangent drawn from the point $(1,4)$ to the parabola $y^2=4x$ is

$\begin{array}{1 1}(A)\;\frac{\pi}{2} \\(B)\;\frac{\pi}{3} \\(C)\;\frac{\pi}{4} \\(D)\;\frac{\pi}{6} \end{array}$

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1 Answer

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Any tangent to $y^2=4x$ is $y=mx +\large\frac{1}{m}$ It passes through $(1,4)$
$4=m+\large\frac{1}{m}$
=> $m^2-4m-1=0$
=> $m_1+m_2 =4 m_1m_2 =1$
$\therefore |m_1-m_2|= 2 \sqrt 3$
$\tan \theta =\large\frac{2 \sqrt 3}{1+1}$
$\qquad=\sqrt 3$
$\theta= \large\frac{\pi}{3}$
Hence B is the correct anwer.
answered Apr 10, 2014 by meena.p
 

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