The Parabola

Focus, Directrix, and the Standard Equation

Key Point (Definition): A parabola is the set of all points equidistant from a fixed point $F$ (the focus) and a fixed line (the directrix): $PF = PB$ for every point $P$, where $PB$ is the perpendicular distance to the directrix.

The axis is the line through $F$ perpendicular to the directrix; the vertex is where the parabola crosses its axis — the midpoint of the focus and the directrix. (If the fixed point lies on the fixed line, the "parabola" degenerates to a straight line through the point, perpendicular to the line.)

Deriving $y^2 = 4ax$. Put the vertex at the origin with focus $F(a, 0)$, $a > 0$, and directrix $x = -a$. For $P(x, y)$, the condition $PF = PB$ reads

$$\sqrt{(x - a)^2 + y^2} = \sqrt{(x + a)^2}$$

Squaring and cancelling $x^2 + a^2$ from both sides leaves $-2ax + y^2 = 2ax$, i.e.

$$y^2 = 4ax$$

The same construction in the other three orientations gives the full family:

Key Point (The four standard parabolas): $y^2 = 4ax$ (opens right, focus $(a,0)$, directrix $x=-a$); $y^2 = -4ax$ (opens left, focus $(-a,0)$, directrix $x=a$); $x^2 = 4ay$ (opens up, focus $(0,a)$, directrix $y=-a$); $x^2 = -4ay$ (opens down, focus $(0,-a)$, directrix $y=a$).

Reading the orientation: a $y^2$ term means the axis is the $x$-axis; an $x^2$ term means the axis is the $y$-axis. The sign of the linear term says which way the curve opens.

Latus Rectum and Working the Standard Forms

Key Point (Latus rectum): The latus rectum is the chord through the focus perpendicular to the axis. For $y^2 = 4ax$ its ends are $(a, \pm 2a)$ and its length is $4a$.

Why: the end $A$ of the latus rectum is equidistant from focus and directrix, and its directrix distance is exactly $2a$ (from $x = -a$ to $x = a$), so $AF = 2a$ on each side of the axis.

The two exam directions:

Equation → data: for $y^2 = 8x$, compare with $y^2 = 4ax$: $a = 2$. Focus $(2, 0)$, axis the $x$-axis, directrix $x = -2$, latus rectum 8.

Data → equation: identify the orientation first, then find $a$.

Focus $(2,0)$ and directrix $x = -2$: opens right, $a = 2$, so $y^2 = 8x$. Vertex $(0,0)$ and focus $(0,2)$: axis is the $y$-axis, opens up, $x^2 = 8y$. Symmetric about the $y$-axis through $(2, -3)$: the point sits below the vertex, so it opens down, $x^2 = -4ay$ with $4 = 12a$, giving $3x^2 = -4y$.

[Board Tip] When a parabola is given by a point it passes through plus its axis, substitute the point into the correctly-oriented form — choosing $y^2 = 4ax$ when the curve actually opens left (or down) produces a negative $a$ and a lost mark. Check which quadrant the given point occupies first.

Solved Examples

Example 1: Reading everything from the equation

Find the focus, axis, directrix and latus rectum of $y^2 = 8x$.

Solution:

Step 1 — Identify the orientation. A $y^2$ term with a positive $x$-side: opens right along the $x$-axis.

Step 2 — Find $a$. Compare with $y^2 = 4ax$: $4a = 8$, so $a = 2$.

Step 3 — Read off the anatomy. Focus $(2, 0)$; axis: the $x$-axis; directrix $x = -2$; latus rectum $4a = 8$.

Takeaway: One comparison gives $a$; every other quantity is a formula in $a$.

Example 2: The same reading, three ways

Find the focus, directrix and latus rectum of: (i) $y^2 = 12x$; (ii) $x^2 = 6y$; (iii) $y^2 = -8x$.

Solution:

Step 1 — (i). $a = 3$: focus $(3, 0)$, directrix $x = -3$, latus rectum 12.

Step 2 — (ii). $x^2$ term, opens up, $a = \frac{3}{2}$: focus $\left(0, \frac{3}{2}\right)$, directrix $y = -\frac{3}{2}$, latus rectum 6.

Step 3 — (iii). Opens left, $a = 2$: focus $(-2, 0)$, directrix $x = 2$, latus rectum 8.

Takeaway: The squared variable names the axis; the sign names the direction; $|4a|$ is always the latus rectum.

Example 3: Downward and leftward flavours

Find the focus, directrix and latus rectum of: (i) $x^2 = -16y$; (ii) $y^2 = 10x$; (iii) $x^2 = -9y$.

Solution:

Step 1 — (i). Opens down, $a = 4$: focus $(0, -4)$, directrix $y = 4$, latus rectum 16.

Step 2 — (ii). Opens right, $a = \frac{5}{2}$: focus $\left(\frac{5}{2}, 0\right)$, directrix $x = -\frac{5}{2}$, latus rectum 10.

Step 3 — (iii). Opens down, $a = \frac{9}{4}$: focus $\left(0, -\frac{9}{4}\right)$, directrix $y = \frac{9}{4}$, latus rectum 9.

Takeaway: Odd coefficients just make $a$ fractional — the reading routine never changes.

Example 4: From focus and directrix

Find the parabola with (i) focus $(2, 0)$ and directrix $x = -2$; (ii) focus $(6, 0)$ and directrix $x = -6$.

Solution:

Step 1 — (i) Orientation. Focus on the positive $x$-axis with the directrix behind the origin: opens right.

Step 2 — (i) Find $a$ and write. $a = 2$: $y^2 = 8x$.

Step 3 — (ii) Same shape. $a = 6$: $y^2 = 24x$.

Takeaway: The vertex is the midpoint of focus and directrix — here the origin, confirming the standard form applies.

Example 5: A downward parabola

Find the parabola with focus $(0, -3)$ and directrix $y = 3$.

Solution:

Step 1 — Orientation. Focus below the origin, directrix above: opens down, form $x^2 = -4ay$.

Step 2 — Find $a$. Focus $(0, -a)$ gives $a = 3$.

Step 3 — Write. $x^2 = -12y$.

Takeaway: The parabola always bends AWAY from its directrix and wraps around its focus.

Example 6: From vertex and focus

Find the parabola with vertex $(0,0)$ and focus at (i) $(0, 2)$; (ii) $(3, 0)$; (iii) $(-2, 0)$.

Solution:

Step 1 — (i). Focus on the positive $y$-axis: $x^2 = 4ay$ with $a = 2$: $x^2 = 8y$.

Step 2 — (ii). Focus on the positive $x$-axis: $y^2 = 12x$.

Step 3 — (iii). Focus on the negative $x$-axis: opens left: $y^2 = -8x$.

Takeaway: Vertex-to-focus direction IS the opening direction — read it straight off the coordinates.

Example 7: Orientation from a point

Find the parabola symmetric about the $y$-axis, vertex at the origin, passing through $(2, -3)$.

Solution:

Step 1 — Choose the correct form. Symmetric about the $y$-axis with the point BELOW the vertex: opens down, $x^2 = -4ay$.

Step 2 — Substitute the point. $2^2 = -4a(-3) = 12a$ gives $a = \frac{1}{3}$.

Step 3 — Write. $x^2 = -\frac{4}{3}y$, i.e. $3x^2 = -4y$.

Takeaway: Locate the point's quadrant BEFORE picking the form — a negative $a$ later means you chose wrong.

Example 8: Axis along the $x$-axis

Find the parabola with vertex $(0,0)$, axis along the $x$-axis, passing through $(2, 3)$.

Solution:

Step 1 — Choose the form. The point $(2, 3)$ has $x > 0$: opens right, $y^2 = 4ax$.

Step 2 — Substitute. $9 = 4a(2) = 8a$: $a = \frac{9}{8}$.

Step 3 — Write. $y^2 = \frac{9}{2}x$, i.e. $2y^2 = 9x$.

Takeaway: The point plugs straight into the correctly-oriented form — one equation, one unknown.

Example 9: Symmetric about the $y$-axis

Find the parabola with vertex $(0,0)$, symmetric about the $y$-axis, passing through $(5, 2)$.

Solution:

Step 1 — Choose the form. Point above the vertex: opens up, $x^2 = 4ay$.

Step 2 — Substitute. $25 = 4a(2) = 8a$: $a = \frac{25}{8}$.

Step 3 — Write. $x^2 = \frac{25}{2}y$, i.e. $2x^2 = 25y$.

Takeaway: Check at the end: $2(25) = 50 = 25(2)$ ✓ — substitution back is free insurance.

Example 10: Latus rectum ends — a coordinate check

For $y^2 = 12x$, verify that the ends of the latus rectum are $(3, \pm 6)$ and that its length is 12.

Solution:

Step 1 — Locate the focal chord. $a = 3$: the latus rectum is the vertical chord at $x = 3$.

Step 2 — Intersect with the parabola. $y^2 = 36$: $y = \pm 6$ — ends $(3, 6)$ and $(3, -6)$.

Step 3 — Measure. Length $= 6 - (-6) = 12 = 4a$. ∎

Takeaway: The ends $(a, \pm 2a)$ and length $4a$ are worth memorising — they anchor countless focal-chord problems.