Answers to Tutorials on the Equation of a Parabola

The answers to the tutorial questions from Equation of a Parabola are presented below with clear explanations and analytic justification.

Question 2

Keeping the values of \( a \), \( h \), and \( k \) fixed, find the equation of the directrix, the coordinates of the vertex \( V \) and focus \( F \), and the equation of the axis of symmetry.

Answer:

• Directrix: \( x = -1 \)
• Vertex: \( V(0, 0) \)
• Focus: \( F(1, 0) \)
• Axis of symmetry: \( y = 0 \)

Question 3

Set \( a = 2 \) and answer the same questions as in Question 2.

Answer:

• Directrix: \( x = -2 \)
• Vertex: \( V(0, 0) \)
• Focus: \( F(2, 0) \)
• Axis of symmetry: \( y = 0 \)

Question 4

Set \( a = 1 \), \( h = 0 \), and vary \( k \). Describe how \( k \) affects the vertex, focus, and axis of symmetry.

Answer:

• Y-coordinate of focus: \( k \)
• Y-coordinate of vertex: \( k \)
• Axis of symmetry: \( y = k \)

Yes, the position of the vertex changes as \( k \) changes.

Question 5

Set \( a = 1 \), \( k = 0 \), and vary \( h \). Describe how \( h \) affects the focus, vertex, and directrix.

Answer:

• X-coordinate of focus: \( h \)
• X-coordinate of vertex: \( h \)
• Directrix: \( x = h - 1 \)

The axis of symmetry does not change position as \( h \) varies.

Question 6

Express the vertex, focus, directrix, and axis of symmetry in terms of \( h \) and \( k \).

Answer:

• Vertex: \( V(h, k) \)
• Focus: \( F(h + 1, k) \)
• Directrix: \( x = h - 1 \)
• Axis of symmetry: \( y = k \)

Question 7

For \( a = 1 \) and \( k = 0 \), determine the number of y-intercepts depending on \( h \).

Answer:

• Two y-intercepts if \( h < 0 \)
• One y-intercept if \( h = 0 \)
• No y-intercepts if \( h > 0 \)

Analytical explanation:

Starting from the standard equation \[ (y - k)^2 = 4a(x - h), \] set \( x = 0 \):

• If \( h < 0 \): two real solutions
• If \( h = 0 \): one real solution
• If \( h > 0 \): no real solutions

Question 8

Explain why this parabola has exactly one x-intercept.

Answer:

Set \( y = 0 \) in \[ (y - k)^2 = 4a(x - h) \] to obtain \[ k^2 = 4a(x - h). \]

This linear equation in \( x \) has exactly one solution: \[ x = h + \frac{k^2}{4a}. \]

Question 9

Show that the equation \[ y^2 - 4y - 4x = 0 \] can be written in standard parabola form.

Solution:

Complete the square:

Thus:

• \( a = 1 \)
• \( h = -1 \)
• \( k = 2 \)

• Vertex: \( V(-1, 2) \)
• Focus: \( F(0, 2) \)
• Axis of symmetry: \( y = 2 \)
• Directrix: \( x = -2 \)