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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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Answer» 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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