Vertex, focus, directrix, and axis of symmetry for horizontal parabolas — with analytical justifications
These are the detailed, step-by-step solutions to the tutorial questions in the Equation of a Parabola module. Each solution shows how the parameters \( a \), \( h \), and \( k \) affect the graph, with analytical justifications that you can replicate on similar problems.
Standard horizontal parabola:
\[ (y - k)^2 = 4a(x - h) \]In this form, \( a \) is the focal parameter — the directed distance from the vertex to the focus. It is not the leading coefficient of a vertical parabola \( y = a(x - h)^2 + k \).
With \( a = 1 \), \( h = 0 \), and \( k = 0 \), find the directrix, the vertex \( V \), the focus \( F \), and the axis of symmetry.
With these values the standard form reduces to:
\[ y^2 = 4x \]so the properties are:
Set \( a = 2 \) (still \( h = 0 \), \( k = 0 \)) and answer the same questions as in Question 2.
The equation becomes:
\[ y^2 = 8x \]so the properties are:
Doubling \( a \) moves both the focus and the directrix twice as far from the vertex, widening the parabola.
Keeping \( a = 1 \) and \( h = 0 \) fixed, vary \( k \). How does \( k \) affect the vertex, the focus, and the axis of symmetry?
Conclusion: Increasing \( k \) shifts the vertex and the focus vertically upward, and shifts the horizontal axis of symmetry upward by the same amount. The directrix (a vertical line) is unaffected because it depends only on \( h \) and \( a \).
Keeping \( a = 1 \) and \( k = 0 \) fixed, vary \( h \). How does \( h \) affect the vertex, the focus, and the directrix?
Conclusion: Varying \( h \) shifts the entire parabola horizontally — vertex, focus, and directrix all translate by \( h \). The axis of symmetry stays horizontal (the line \( y = 0 \)).
Express the vertex, focus, directrix, and axis of symmetry in terms of \( h \) and \( k \) (assuming \( a = 1 \)).
For \( a = 1 \) and \( k = 0 \), determine how the number of y-intercepts depends on \( h \).
Set \( x = 0 \) in \( (y - k)^2 = 4a(x - h) \). With \( a = 1 \) and \( k = 0 \):
\[ y^2 = -4h \]The number of real solutions depends on the sign of \( -4h \):
Explain why this standard horizontal parabola has exactly one x-intercept.
Set \( y = 0 \) in the standard equation:
\[ (0 - k)^2 = 4a(x - h) \] \[ k^2 = 4a(x - h) \]For \( a \ne 0 \), this is a linear equation in \( x \) — so it has exactly one solution:
\[ x = h + \frac{k^2}{4a} \]Geometrically, any horizontal parabola opens purely left or right, so it crosses the x-axis only once.
Show that \( y^2 - 4y - 4x = 0 \) can be written in standard parabola form, and find its properties.
Isolate the \( y \)-terms:
\[ y^2 - 4y = 4x \]Complete the square by adding \( 4 \) to both sides:
\[ y^2 - 4y + 4 = 4x + 4 \] \[ (y - 2)^2 = 4(x + 1) \]Comparing with \( (y - k)^2 = 4a(x - h) \):
Since \( a > 0 \), the parabola opens to the right.
Properties:
| Property | Formula |
|---|---|
| Standard form | \( (y - k)^2 = 4a(x - h) \) |
| Vertex | \( V(h, k) \) |
| Focus | \( F(h + a, k) \) |
| Directrix | \( x = h - a \) |
| Axis of symmetry | \( y = k \) |
| x-intercept | \( x = h + \dfrac{k^2}{4a} \) |
| y-intercepts | \( y = k \pm \sqrt{-4a(x_0 - h)} \) evaluated at \( x_0 = 0 \) — two, one, or none depending on \( -4ah \) |
\( (y - k)^2 = 4a(x - h) \), where \( V(h, k) \) is the vertex, \( F(h + a, k) \) is the focus, \( x = h - a \) is the directrix, and \( y = k \) is the axis of symmetry.
From the standard form, the focus is \( F(h + a, k) \) — displaced horizontally from the vertex by the focal parameter \( a \).
The directrix is the vertical line \( x = h - a \).
Setting \( y = 0 \) gives \( k^2 = 4a(x - h) \), which is linear in \( x \) (as long as \( a \ne 0 \)), so there is exactly one solution: \( x = h + k^2/(4a) \).
Setting \( x = 0 \) gives \( (y - k)^2 = -4ah \). There are two y-intercepts if \( -4ah > 0 \), one if \( -4ah = 0 \), and none if \( -4ah < 0 \).
Last updated: October 17, 2026