Question:
2  Keep the values of a, h and k as above (do not change the positions of the sliders). Find the equation of the directrix and the coordinates of the vertex V and focus F. Find the equation of the axis of symmetry of the parabola (line through V and F).
Answer:
directrix: vertical (green) line: x = 1
Vertex V: (0 , 0)
Focus F: (1 , 0)
Axis of Parabola line y = 0 (x axis).
Question:
3  Use the top slider to set a = 2 and answer the same questions as in part 2 above.
Answer:
directrix: vertical (green) line: x = 2
Vertex V: (0 , 0)
Focus F: (2 , 0)
Axis of Parabola line y = 0 (x axis).
Question:
4  Set a = 1, h = 0 and change k (using the slider). Find a relationship between the ycoordinate of F and parameter k. Find a relationship between the ycoordinate of V and k. Find a relationship between the position (or equation) of the axis of the parabola and k. Does the position of the vertex change?
Answer:
y coordinate of focus F = k
y coordinate of vertex V = k
equation of axis of parabola y = k.
Yes the position of the vertex change as k changes.
Question:
5  Set a = 1, k = 0 and change h (using the slider). Find a relationship between the xcoordinate of F and parameter h. Find a relationship between the xcoordinate of V and h. Find a relationship between the position (or equation) of the directrix of the parabola and h. Does the position of the axis change?
Answer:
x coordinate of focus F = h
x coordinate of vertex V = h
equation of directrix of parabola x = h  1.
No, the position of the axis of the parabola does not change as h changes.
Question:
6  Use parts 1,2,3,4 and 5 above to find the coordinates of V and F and the equations of the directrix and axis of the parabola in terms of h and k.
Answer:
Vertex V is at the point (h , k).
Focus F is at the point (h + 1 , k)
Equation of directrix: x = h  1
Equation of axis: y = k
Question:
7  Set a = 1, k = 0 and change h. Which values of h give two yintercepts? Which values of h give no yintercepts? Which values of h give one yintercept?Explain your answers analytically.(Hint: find the yintercepts by setting x = 0 and solve for y).
Answer:
two yintercepts when h < 0
one yintercept when h = 0
no yintercept when h > 0
analytical
To find the yintercepts, set x = 0 to zero in the equation (y  k)^{2} = 4a(x  h) (coefficient a is equal to 1) of the parabola and solve for y.
(y  k)^{2} = 4h
If h < 0, the above equation has two real solutions
y = k + sqrt(4h) and y = k  sqrt(4h)
If h = 0, the above equation has one real solution
y = k
If h > 0, the above equation has no real solutions.
Question:
8  Investigate the xintercept. Explain why the parabola as defined above has one xintercept only.
Answer:
To find the x intercept we set y = 0 in the equation (y  k)^{2} = 4a(x  h) and solve for x.
k^{2} = 4a(x  h)
The above equation will always have one solution given by
x = h + k^{2} / 4a
Question:
9  Exercise: Show that the following equation
y^{2}  4y  4x = 0
can be written as
(yk)^{2} = 4a(x  h)
Hint: put all terms with y and y^{2} together in one side and all terms with x in the other side of the equation. Complete the square for the expression containing y and y^{2}.
Find a, h and k. Find the coordinates of V and F. Find the equations of the axis and directrix of this parabola. Put the values of a, h and k in the applet and check your answer.
Answer:
We rewrite the equation as
y^{2}  4y = 4x
complete the square
y^{2}  4y + (4/2)^{2} = 4x + (4/2)^{2}
(y  2)^{2} = 4(x + 1)
hence: a = 1, k = 2 and h = 1
Vertex at V(h , k) = (1 , 2)
Focus at F(h + a, k) = (0 , 2)
Axis, horizontal line, y = k = 2
Directrix, vertical line, x = h  a = 2
