Parabola: focus and directrix
A satellite dish and a car headlight reflect all rays to (or from) a single point. Their cross-section is a parabola.
A parabola is the set of points that are equally far from a fixed point F (the focus) and a fixed line l (the directrix). The line through F perpendicular to l is the axis; the point where the curve cuts the axis is the vertex.
Formula: distance to focus = distance to directrix
In the figure: Parabola y² = 6x: focus (1.5, 0), directrix x = −1.5, vertex at the origin.
Rule 1: Check a point — Distance to F must equal distance to l.
- Easy: F(1, 0), directrix x = −1: point (1, 2): distance to F = 2, to line = 2. On the curve.
- Medium: Point (4, 4): distance to F = 5, to line = 5. On the curve.
- Exam-type: Point (0, 0): both are 1, the vertex.
Rule 2: Parts of a parabola — Vertex is halfway between focus and directrix.
- Easy: Focus (2, 0), directrix x = −2: vertex (0, 0).
- Medium: If the distance from focus to directrix is 2a, the vertex is at distance a from each.
- Exam-type: Axis of y² = 4ax is the x-axis.
Every point on the parabola is as far from the focus as from the directrix.
Measuring the distance to the directrix along a slant instead of perpendicularly.
Focus-and-line, equal distances.
Practice questions
- 1What is the vertex of a parabola with focus (3, 0) and directrix x = −3?
- 2Is (1, 2) on the parabola with focus (1, 0) and directrix x = −1?
- 3What is the axis of a parabola with focus (0, 2) and directrix y = −2?
- 4What is the distance between focus and directrix of y² = 8x?
- 5If the focus is (4, 0), what is the directrix of the parabola with vertex at the origin?
- 6Check if (9, 6) is on the parabola with focus (4, 0) and directrix x = −4.
- 7Where does the parabola cut its axis?
- 8A point is 5 units from the focus. How far is it from the directrix?
Show answers (try them yourself first)
- 1(0, 0)
- 2Distance to focus 2; to directrix 2. Yes.
- 3The y-axis
- 42a = 4, with a = 2
- 5x = −4
- 6Distance to focus: √(25 + 36) = √61. To directrix: 13. Not equal, not on the curve
- 7At the vertex.
- 85 (equal distances)
Standard equations of a parabola
Turn a parabola to face right, left, up or down: its equation changes in a neat way.
With vertex at the origin and a > 0: y² = 4ax opens right (focus (a, 0), directrix x = −a); y² = −4ax opens left; x² = 4ay opens up (focus (0, a), directrix y = −a); x² = −4ay opens down.
Formula: y² = 4ax
Rule 1: Find focus and directrix — Compare with 4a.
- Easy: y² = 8x: a = 2, focus (2, 0), directrix x = −2.
- Medium: x² = 12y: a = 3, focus (0, 3), directrix y = −3.
- Exam-type: y² = −16x: a = 4, focus (−4, 0), directrix x = 4.
Rule 2: Write the equation — Use the focus or directrix to find a.
- Easy: Focus (3, 0): y² = 12x.
- Medium: Focus (0, −2): x² = −8y.
- Exam-type: Directrix x = 5: y² = −20x.
The variable that is squared tells the axis: y² means the axis is the x-axis.
Taking a as the coefficient of x instead of dividing it by 4.
Squared letter = axis. Sign = direction.
Spot the difference
| y² = 4ax | x² = 4ay |
|---|---|
| Axis: x-axis | Axis: y-axis |
| Opens right (left if −) | Opens up (down if −) |
Practice questions
- 1Find the focus of y² = 20x.
- 2Find the directrix of x² = 16y.
- 3Write the parabola with focus (2, 0).
- 4Write the parabola with focus (0, 4).
- 5Write the parabola that opens left with focus (−3, 0).
- 6Write the parabola with vertex at the origin and directrix y = 3.
- 7Find a point on y² = 16x with x-coordinate 4.
- 8A parabola x² = 4ay passes through (4, 2). Find a.
Show answers (try them yourself first)
- 1(5, 0)
- 2y = −4
- 3y² = 8x
- 4x² = 16y
- 5y² = −12x
- 6x² = −12y
- 7y² = 64, y = ±8: (4, 8) or (4, −8)
- 816 = 8a, a = 2
Latus rectum of a parabola
The chord through the focus perpendicular to the axis tells us how wide the parabola is at the focus.
The latus rectum is the chord through the focus perpendicular to the axis. For y² = 4ax its length is 4a, with end points (a, 2a) and (a, −2a).
Formula: length of latus rectum = 4a
Rule 1: Find the length — Find a from the equation.
- Easy: y² = 8x: a = 2, length 8.
- Medium: x² = 12y: a = 3, length 12.
- Exam-type: y² = 5x: a = 5/4, length 5.
Rule 2: End points — Put x = a in y² = 4ax.
- Easy: y² = 8x: ends (2, 4) and (2, −4).
- Medium: y² = 4x: ends (1, 2) and (1, −2).
- Exam-type: x² = 8y: ends (4, 2) and (−4, 2).
Length of latus rectum = coefficient of x (or y) in the standard equation, ignoring the sign.
Taking the latus rectum as 2a.
Latus rectum = the number in front.
Practice questions
- 1Find the length of the latus rectum of y² = 12x.
- 2Find the length of the latus rectum of x² = 20y.
- 3Find the end points of the latus rectum of y² = 16x.
- 4The latus rectum of y² = 4ax has length 10. Find a.
- 5Find the focus and length of latus rectum of 3y² = 24x.
- 6Find the equation of the parabola with vertex at the origin and latus rectum 6 opening right.
- 7Find the area of the triangle formed by the vertex and the ends of the latus rectum of y² = 4x.
- 8Find the end points of the latus rectum of x² = −12y.
Show answers (try them yourself first)
- 112
- 220
- 3(4, 8) and (4, −8)
- 4a = 2.5
- 5y² = 8x: focus (2, 0), length 8
- 6y² = 6x
- 7base 4, height 1: area 2
- 8(6, −3) and (−6, −3)
Parabola problems
Given only a point on the curve and its direction, we can still find the equation.
Steps: decide the axis and direction from the data; write y² = ±4ax or x² = ±4ay; use the given point to find a.
Rule 1: Using a point on the curve — Substitute into the standard form.
- Easy: Opens right through (2, 4): 16 = 4a(2), a = 2: y² = 8x.
- Medium: Opens up through (4, 8): 16 = 4a(8), a = 1/2: x² = 2y.
- Exam-type: Opens left through (−3, 6): 36 = 4a(3), a = 3: y² = −12x.
Rule 2: Applications — Cables, arches, reflectors.
- Easy: A dish 8 cm wide and 2 cm deep: x² = 4ay through (4, 2): a = 2, the focus is 2 cm from the vertex.
- Medium: An arch of height 4 m and width 8 m: x² = −4ay through (4, −4) gives a = 1.
- Exam-type: A bridge cable x² = 40y gives the height 10 m at 20 m from the centre.
Always check by substituting the point back.
Choosing the wrong direction (sign) of the parabola.
Point in, solve for a.
Practice questions
- 1Find the parabola y² = 4ax through (1, 2).
- 2Find the parabola x² = 4ay through (2, 1).
- 3A parabola opens down and passes through (4, −8). Find its equation.
- 4Find the focus of y² = 6x.
- 5A reflector is 12 cm across and 4.5 cm deep. Where is the bulb (focus) placed from the vertex?
- 6A parabola y² = 4ax passes through (3, 6). Find the focus.
- 7A cable follows x² = 8y. Find the height at 4 m from the centre line.
- 8Find the equation of the parabola with focus (6, 0) and directrix x = −6.
Show answers (try them yourself first)
- 1a = 1: y² = 4x
- 2a = 1: x² = 4y
- 3x² = −4ay: 16 = 32a, a = 1/2: x² = −2y
- 4(1.5, 0)
- 5x² = 4ay through (6, 4.5): a = 36/18 = 2 cm
- 636 = 12a, a = 3: focus (3, 0)
- 716 = 8y, y = 2 m
- 8y² = 24x