The coordinates of the focus of the parabola described parametrically by $x = 5t^2 + 2, y = 10t + 4 \text{ arc}$
Step-by-Step Solution
Key Concept: For a parabola $(y-k)^2 = 4a(x-h)$, the focus is at $(h+a, k)$ and the vertex is at $(h, k).
From the parametric equations $x = 5t^2 + 2$ and $y = 10t + 4$, we eliminate $t$. From the second equation, $t = \frac{y-4}{10}$. Substituting into the first: $x - 2 = 5\left(\frac{y-4}{10}\right)^2 = \frac{(y-4)^2}{20}$, which gives $(y-4)^2 = 20(x-2)$. This is a parabola with vertex at $(2, 4)$, opening rightward with $4a = 20$, so $a = 5$. The focus is at $(2 + 5, 4) = (7, 4)$.
Correct Answer: (7, 4)