I'm confused...

Question

There are many points in the first quadrant (of the xy-plane) that are the same distance from the x-axis as they are from the point (0, 2). Make a sketch that shows several of them. Use your ruler to check that the points you plotted satisfy the required equidistance property.

Answer 1

if you don't immediately recognize x-axis as directix and (0,2) as focus, problem wants:

√[(x-0)²+(y-2)²]=√[(x-x)²+(y-0)²]

and wants only Q1 (x,y) pairs.

y=(x²+4)/4

x≥0

x=10, y=26

which is 26 from y=0 x=10

and √(10²+24²) from (0,2)

√(10²+24²)=26

Answer 2

the graph is an upward opening parabola with directrix y=0, focus (0,2), vertex (0,1)= midpoint between focus and directrix

y=a(x^2) + 1

solve for a by plugging in any other point on the parabola, such as (2,2) which is equidistant from the focus and directrix

2 = a(2)^2 +1

4a =1

a = 1/4

y = (1/4)x^2 +1

other points on the parabola include

(-2,2), (3,13/4), (-3,13/4), (4,5,), (-4,5), (5,29/4),(-5,29/4)