Q.299. To find the directrix of the parabola given the parametric point.

Question 299.
x = t^2 + 1, y = 2t + 1.

Answer 299.
Eliminating paramegter t from the two equations,
x = [(y-1)/2]^2 + 1
=> 4(x-1) = (y-1)^2
Shifting origin to (1, 1), the new coordinates (x', y') and old coordinates are related as
x' = x - 1 and y' = y - 1
=> y'^2 = 4x' is the standard equation of the parabola whose directrix is
x' = -1
=> x - 1 = -1
=> x = 0 (which is y-axis) is the directrix of the given ;parabola.

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