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Question

For the curve represented parametrically by the equations, x=2cott+1 & y=tant+cott

A
tangent at t=π4 is parallel to x - axis
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B
normal at t=π4 is parallel to y - axis
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C
tangent at t=π4 is parallel to the line y=x
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D
tangent and normal intersect at the point (2,1)
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Solution

The correct options are
A tangent at t=π4 is parallel to x - axis
B normal at t=π4 is parallel to y - axis
dy=sec2tcosec2t.dt ...(i)
dx=2sec2t.dt ...(ii)

dydx=sec2tcosec2t.dt2sec2t.dt
=sec2tcosec2t2sec2t.
Hence
dydxt=π4 =0 slope of tangent

Slope of normal
=dxdy
=2sec2tsec2tcosec2t
Hence =dxdyt=π4=
Hence the normal is perpendicular to x axis and thus parallel to y axis.
and tangent is parallel to x - axis.

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