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Question

If the tangent to y2=4ax at the point (at2,2at) where |t|>1 is a normal to x2y2=a2 at the point (asecθ,atanθ), then 
  1. t=cosec θ
  2. t=secθ
  3. t=2tanθ
  4. t=2cotθ


Solution

The correct options are
A t=cosec θ
C t=2tanθ
Equation 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 x2y2=a2 :
dxdy=yx
[dxdy](asecθ,atanθ)=atanθasecθ=sinθ

Therefore, equation of the normal is 
yatanθ=sinθ(xasecθ)
y=xcosec θ+2atanθ     (2)

Comparing (1) and (2), we get 
t=cosec θ and t=2tanθ

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