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The parametric equation of a parabola is...

The parametric equation of a parabola is `x=t^2+1,y=2t+1.` Then find the equation of the directrix.

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We have `x=t^(2)+1,y=2t+1`.
Eliminating t, we get
`x-1=((y-1)/(2))^(2)`
`or" "(y-1)^(2)=4(x-1)`
Comparing with `(y-k)^(2)=4a(x-h)`, we have vertex (1,1) and axis parallel to x-axis having equation y-1=0.
Also, 4a=4. So, a = 1
Directrix is parallel to y-axis and lies to the left of the vertex at distance a units from it.
`So, equation of directrix is x=1-1 or x=0.
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