Question

Any point on the parabola whose focus is (0, 1) and the directrix is $x + 2 = 0$ is given by

• A. (t^2 + 1, 2t - 1)
• B. (t^2 + 1, 2t + 1)
• C. (t^2, 2t)
• D. (t^2 - 1, 2t + 1)

Answer 1

Answer: (D)

(D)

Answer 2

Let $P(x, y)$ be any point on the parabola. The distance from $P$ to the focus $F(0, 1)$ must equal the distance from $P$ to the directrix $x+2=0$. The squared distance to the focus is $x^2 + (y-1)^2$. The squared distance to the directrix is $(x+2)^2$. Equating these gives $x^2 + (y-1)^2 = (x+2)^2$, which simplifies to $y^2 - 2y - 3 = 4x$. Substituting option (D), $(t^2 - 1, 2t + 1)$, into the equation: $4(t^2 - 1) = (2t + 1)^2 - 2(2t + 1) - 3$ $4t^2 - 4 = (4t^2 + 4t + 1) - (4t + 2) - 3$ $4t^2 - 4 = 4t^2 - 4$. This is true for all $t$.