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Question

If the tangent to y2=4ax at the point (at2,2at) where |t|>1 is a normal to x2−y2=a2 at the point (asecθ,atanθ), then

A
t=cosec θ
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B
t=secθ
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C
t=2tanθ
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D
t=2cotθ
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Solution

The correct options are A t=−cosec θ C t=2tanθEquation of tangent y2=4ax at the point (at2,2at) is 2aty=2a(x+at2) ⇒ty=x+at2 ⇒y=xt+at ⋯(1) Equation of normal to x2−y2=a2 : dxdy=yx [−dxdy](asecθ,atanθ)=−atanθasecθ=−sinθ Therefore, equation of the normal is y−atanθ=−sinθ(x−asecθ) ⇒y=−xcosec θ+2atanθ ⋯(2) Comparing (1) and (2), we get t=−cosec θ and t=2tanθ

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