The anwsers to the tutorial in Equation of Parabola are presented.

## Question2 - 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). ## Question3 - 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). ## Question4 - Set a = 1, h = 0 and change k (using the slider). Find a relationship between the y-coordinate of F and parameter k. Find a relationship between the y-coordinate 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. ## Question5 - Set a = 1, k = 0 and change h (using the slider). Find a relationship between the x-coordinate of F and parameter h. Find a relationship between the x-coordinate 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. ## Question6 - 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 ## Question7 - Set a = 1, k = 0 and change h. Which values of h give two y-intercepts? Which values of h give no y-intercepts? Which values of h give one y-intercept?Explain your answers analytically.(Hint: find the y-intercepts by setting x = 0 and solve for y).Answer
two y-intercepts when h < 0 one y-intercept when h = 0 no y-intercept when h > 0 analytical
To find the y-intercepts, 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 - √(-4h) If h = 0, the above equation has one real solution y = k If h > 0, the above equation has no real solutions. ## Question8 - Investigate the x-intercept. Explain why the parabola as defined above has one x-intercept 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
## Question9 -Exercise: Show that the following equation
^{2} - 4y - 4x = 0can be written as ^{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 |