Question
Any point on the parabola whose focus is $$\left( {0,\,1} \right)$$ and the directrix is $$x + 2 = 0$$ is given by :
A.
$$\left( {{t^2} + 1,\,2t - 1} \right)$$
B.
$$\left( {{t^2} + 1,\,2t + 1} \right)$$
C.
$$\left( {{t^2},\,2t} \right)$$
D.
$$\left( {{t^2} - 1,\,2t + 1} \right)$$
Answer :
$$\left( {{t^2} - 1,\,2t + 1} \right)$$
Solution :
The equation is $${\left( {x - 0} \right)^2} + {\left( {y - 1} \right)^2} = {\left( {\frac{{x + 2}}{{\sqrt 1 }}} \right)^2}$$
$${\text{or }}{\left( {y - 1} \right)^2} = 4\left( {x + 1} \right)$$
Clearly, $$x = {t^2} - 1$$ and $$y = 2t + 1$$ satisfy it for all t.