Conic Sections · Lesson 2 of 5
Parabolas: Focus, Directrix, and Orientation
“Build a parabola from equal distances and use its orientation, focus and focal chord to solve problems.”
• Explain the focus–directrix condition, axis, vertex and degenerate case of a parabola. • Derive the origin-centred standard equation and verify its converse. • Distinguish all four orientations from the squared variable and the sign of the other term. • Derive the latus rectum length and find its endpoints. • Construct a standard parabola from a focus, directrix, vertex or point condition.
Equal distances to a point and a line
A circle fixes the distance to one point. A parabola uses a different rule: its points are equally far from a fixed point and a fixed straight line. The curve bends because moving away from the point also changes the distance to the line. The important comparison is between two distances, not between a distance and a coordinate.
A parabola is the set of points in a plane equidistant from a fixed point, the focus, and a fixed line, the directrix. For an ordinary parabola the focus does not lie on the directrix. Distance to a line always means perpendicular distance.
Draw a perpendicular from the focus to the directrix. The line containing this perpendicular is the axis of the parabola. On this axis, the midpoint between the focus and the directrix is equally far from each, so it belongs to the curve. This point is the vertex. It is where the parabola meets its axis and turns back symmetrically.
The axis is the line through the focus perpendicular to the directrix. The vertex is the point where the parabola meets this axis; it is midway between the focus and the foot of the perpendicular from the focus to the directrix.
If the focus lies on the directrix, the locus collapses to a straight line perpendicular to the directrix through that point. To see why, choose focus (0, 0) and directrix x = 0. Equal distances give √(x² + y²) = |x|, so y² = 0. Thus y = 0 is the degenerate locus. Keeping the focus off the directrix avoids this collapse.
Deriving a right-opening parabola
Place the vertex at the origin and the axis on the x-axis. Choose focus F(a, 0) with a > 0 and directrix x = −a. The focus and directrix are 2a apart, while the vertex is a from each. This arrangement turns the geometric condition into a manageable equation.
For an arbitrary point P(x, y), the perpendicular foot on the directrix is B(−a, y). Therefore PF = √((x − a)² + y²), while PB = |x + a|. We keep the absolute value because distance cannot be negative. Equating the distances and squaring gives (x − a)² + y² = (x + a)².
Expand both squares: x² − 2ax + a² + y² = x² + 2ax + a². Cancel x² and a², then move the remaining x terms to obtain y² = 4ax. The factor 4 appears because the two cross terms differ by 4ax; it is not an arbitrary convention.
To prove the converse, suppose P satisfies y² = 4ax. Then PF² = (x − a)² + 4ax = (x + a)², so PF = |x + a| = PB. Thus every solution of the equation has the required equal distances. Also, y² ≥ 0 and a > 0 imply x ≥ 0. On this curve |x + a| = x + a, and the curve occupies the first and fourth quadrants together with its vertex.
The axis is the entire x-axis as a line; the curve opens in the positive x direction. Replacing y by −y does not change the equation, so the points (x, y) and (x, −y) occur together. This proves symmetry about the x-axis. It does not imply symmetry about the y-axis.
Problem
Verify that P(2, 4) lies on y² = 8x and is equally far from its focus and directrix.
- 1.Here 4a = 8, so a = 2, F = (2, 0) and the directrix is x = −2.
- 2.Substitution gives 4² = 8 × 2 = 16, so P lies on the curve.
- 3.PF = √((2 − 2)² + 4²) = 4. The perpendicular distance to x = −2 is |2 + 2| = 4. The equation and the geometric test agree.
All four orientations
Rotating or reflecting the same equal-distance arrangement produces the other standard forms. Keep a positive in every row; use the sign of the linear term to record the direction. The variable that is squared measures displacement across the axis, while the variable that is not squared measures movement along the axis.
| Equation, a > 0 | Opening and axis | Focus | Directrix |
|---|---|---|---|
| y² = 4ax | Right; y = 0 | (a, 0) | x = −a |
| y² = −4ax | Left; y = 0 | (−a, 0) | x = a |
| x² = 4ay | Up; x = 0 | (0, a) | y = −a |
| x² = −4ay | Down; x = 0 | (0, −a) | y = a |
For the left-opening equation, y² ≥ 0 requires x ≤ 0. For upward opening, x² ≥ 0 requires y ≥ 0; for downward opening it requires y ≤ 0. A y² equation is symmetric about the x-axis, while an x² equation is symmetric about the y-axis. The focus always sits on the opening side of the vertex, with the directrix equally far away on the opposite side.
In y² = 12x the axis is the x-axis. Here y changes to ±y without altering the equation. A negative linear coefficient changes the opening direction; it does not make the positive distance a negative.
We use vertex-at-origin forms in this lesson. Knowing only that a parabola is symmetric about the y-axis does not force its vertex to be the origin: it could have vertex elsewhere on that axis. When constructing an equation from a point, the origin-vertex condition must be stated. We will make that assumption explicit rather than quietly add it.
The latus rectum: a chord through the focus
A chord is a segment joining two points on a curve. The latus rectum is the special chord through the focus perpendicular to the axis. Its length provides a useful size measure and is easy to calculate from the standard equation.
The latus rectum is the line segment through the focus, perpendicular to the axis, with both endpoints on the parabola.
For y² = 4ax, the focus has x-coordinate a, so the latus rectum lies on x = a. Substitute into the equation: y² = 4a², giving y = ±2a. Thus its endpoints are (a, 2a) and (a, −2a); its full length is 2a − (−2a) = 4a. Each half has length 2a.
There is also a geometric explanation. An endpoint is 2a from the directrix x = −a, because it lies on x = a. Its distance from the focus must therefore also be 2a. The focus bisects the chord by symmetry, giving total length 4a. This connects the formula directly to equal distances.
Problem
Find the focus, axis, directrix, latus rectum length and endpoints of y² = 8x.
- 1.Compare with y² = 4ax: 4a = 8, giving a = 2. The positive coefficient means opening to the right.
- 2.The focus is (2, 0), the axis is y = 0 and the directrix is x = −2.
- 3.The latus rectum has length 4a = 8. On x = 2, y² = 16, so its endpoints are (2, 4) and (2, −4).
Problem
Find the geometric features of x² = −16y.
- 1.Compare with x² = −4ay, keeping a positive: 4a = 16 and a = 4.
- 2.The vertex is (0, 0), the axis is x = 0, the focus is (0, −4) and the directrix is y = 4.
- 3.The latus rectum lies on y = −4. Then x² = 64, so endpoints are (−8, −4) and (8, −4), with total length 16.
- 4.The curve requires y ≤ 0, consistent with downward opening.
Constructing an equation from geometry
Start by locating the axis and opening direction, then find a. A supplied focus and directrix determine the vertex as a midpoint. If the vertex is already given as the origin, a is simply the positive distance from that vertex to the focus. A point on the curve can then determine the remaining coefficient by substitution.
Problem
Find the equation for focus (2, 0) and directrix x = −2.
- 1.The axis is horizontal. The midpoint between the focus and its foot (−2, 0) on the directrix is (0, 0), so the vertex is the origin.
- 2.The focus is on the positive x side and its distance from the vertex is 2. Use y² = 4ax with a = 2.
- 3.Therefore y² = 8x. Both the focus and directrix match the standard-form features.
Problem
Find the equation for vertex (0, 0) and focus (0, 2).
- 1.The axis is the y-axis and the focus is above the vertex, so choose x² = 4ay.
- 2.The distance a is 2, giving x² = 8y.
- 3.The implied directrix is y = −2, equally far below the vertex. This checks the geometry.
Problem
A parabola has vertex (0, 0), is symmetric about the y-axis, and passes through (2, −3). Find its equation.
- 1.A y-axis parabola has a standard equation x² = Ky. The negative y-coordinate with x ≠ 0 forces K < 0.
- 2.Write x² = −4ay with a > 0. Substitute the point: 4 = −4a(−3) = 12a, so a = 1/3.
- 3.The equation is x² = −(4/3)y, equivalently 3x² = −4y.
- 4.The focus is (0, −1/3) and directrix y = 1/3. Its latus rectum has length 4/3. Without the vertex condition, symmetry and one point would not determine a unique parabola.
Problem
Find the parabola with vertex (0, 0), axis along the x-axis, passing through (2, 3).
- 1.The positive x-coordinate places the nonvertex point on the right, so use y² = 4ax.
- 2.Substitute: 9 = 8a, giving a = 9/8.
- 3.Thus y² = (9/2)x. The focus is (9/8, 0), and directrix x = −9/8. Fractional focal distance is entirely valid.
State the vertex and axis when using a standard form. A point condition only determines the remaining coefficient after the form has been chosen. Also measure distance to the directrix perpendicularly; an oblique segment gives the wrong condition.
Check your understanding
Use the squared coordinate to identify the axis, then use the sign to identify direction. For calculation questions, extract a from the whole coefficient 4a before writing the focus or directrix.
Quiz
What does the defining condition of a parabola compare?
What is the focus of y² = −12x?
What is the directrix of x² = 8y?
Which equation opens downward?
What is the latus rectum length of y² = 10x?
A standard parabola with vertex at the origin has focus (0, −3). Which equation is correct?
For y² = 4ax with a > 0, which pair gives the latus rectum endpoints?
Practice Problems
- Derive y² = 4ax from focus (a, 0) and directrix x = −a, then verify the converse without dropping the absolute value prematurely.
- Find the focus, directrix, axis, latus rectum length and endpoints for y² = −8x and x² = 6y.
- Construct the parabola with focus (0, −3) and directrix y = 3, and explain why its vertex is the origin.
- Find the standard parabola with vertex (0, 0), axis along the y-axis and passing through (5, 2).
- Explain why symmetry about the y-axis plus a single point is insufficient unless a vertex condition is also supplied.
- For a right-opening standard parabola, show geometrically and algebraically why each half of its latus rectum is 2a.
- Explain the degenerate locus when the focus lies on its directrix, using a coordinate argument.
Key Takeaways
• A parabola equates distance to a focus with perpendicular distance to a directrix. • The vertex lies midway between the focus and directrix along the axis; a is a positive distance. • The standard forms are y² = ±4ax and x² = ±4ay, with axis and opening determined by the variables and sign. • The latus rectum passes through the focus perpendicular to the axis and has full length 4a. • State the origin-vertex assumption when a point condition is used to determine a standard parabola.