### Instruction: A parabola has vertex $V = (0,0)$ and focus $F = (0,1).$  Let $P$ be a point in the first quadrant, lying on the parabola, so that $PF = 101.$  Find $P.$.### Response: I know that a parabola is the set of points that are equidistant from a focus and a directrix, which is a line perpendicular to the axis of symmetry of the parabola.
In this case, the axis of symmetry is the $y$-axis, and the directrix is the line $y = -1.$
So, I need to find a point $P = (x,y)$ that satisfies $PF = 101$ and $P$ is also $101$ units away from the directrix.
I can use the distance formula to write two equations for $P.$
The first one is $PF^2 = 101^2,$ which expands to $(x - 0)^2 + (y - 1)^2 = 101^2.$
The second one is $P$'s distance to the directrix squared, which is $(y + 1)^2 = 101^2.$
I can simplify the second equation by taking the square root of both sides and getting $y + 1 = \pm 101.$
Since $P$ is in the first quadrant, I can ignore the negative sign and take $y + 1 = 101.$
This means $y = 100.$
Now, I can plug this value of $y$ into the first equation and solve for $x.$
I get $(x - 0)^2 + (100 - 1)^2 = 101^2.$
Simplifying, I get $x^2 + 9801 = 10201.$
Subtracting 9801 from both sides, I get $x^2 = 400.$
Taking the square root of both sides, I get $x = \pm 20.$
Again, since $P$ is in the first quadrant, I can ignore the negative sign and take $x = 20.$
Therefore, the point $P$ is $(20, 100).$