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The parabola

Covers AS91573

What defines a parabola

A parabola is the set of points that are the same distance from a fixed point (the focus) as from a fixed line (the directrix).
That definition is what makes parabolas useful. Every ray arriving parallel to the axis reflects off the curve and passes through the focus — which is why satellite dishes, headlight reflectors and solar concentrators are parabolic.
A parabola is unbounded: it opens out forever, unlike a circle or ellipse.

The two standard forms

Opening left or right (axis horizontal):

$(y - k)^2 = 4p(x - h)$

Opening up or down (axis vertical):

$(x - h)^2 = 4p(y - k)$

Each part:
- $(h, k)$ — the vertex, the turning point of the curve
- $p$ — the focal length: the distance from the vertex to the focus, and from the vertex to the directrix
- The squared variable tells you the axis. $y$ squared → horizontal axis; $x$ squared → vertical axis.
The sign of $p$ gives the direction of opening:

Form	$p > 0$	$p < 0$
$(y-k)^2 = 4p(x-h)$	opens right	opens left
$(x-h)^2 = 4p(y-k)$	opens up	opens down

Locating the focus and directrix

The parabola opens towards the focus and away from the directrix, and the vertex sits exactly halfway between them.

Form	Vertex	Focus	Directrix
$(y-k)^2 = 4p(x-h)$	$(h, k)$	$(h + p,\ k)$	$x = h - p$
$(x-h)^2 = 4p(y-k)$	$(h, k)$	$(h,\ k + p)$	$y = k - p$

Worked through — $y^2 = 12x$ $y^{2} = 12 x$ :
- Compare with $(y-k)^2 = 4p(x-h)$ : $h = 0$ , $k = 0$ , $4p = 12$ so $p = 3$
- Vertex $(0,0)$ , focus $(3, 0)$ , directrix $x = -3$ , opening right

The latus rectum

The latus rectum is the chord through the focus perpendicular to the axis.
Its length is $|4p|$ — exactly the coefficient in the standard form.
It is the fastest way to sketch the width of the curve. For $y^2 = 12x$ the parabola is 12 units wide at the focus, so it passes through $(3, 6)$ and $(3, -6)$ .

From the $y = ax^2 + bx + c$ you already know

The Level 2 form $y = ax^2 + bx + c$ $y = a x^{2} + b x + c$ is a parabola with a vertical axis. Completing the square converts it:
- $y = 2x^2 - 12x + 23$
- $y = 2(x^2 - 6x) + 23 = 2(x - 3)^2 - 18 + 23 = 2(x-3)^2 + 5$
- Rearranged into conic form: $(x - 3)^2 = \tfrac{1}{2}(y - 5)$
- So $4p = \tfrac{1}{2}$ , giving $p = \tfrac{1}{8}$ : vertex $(3, 5)$ , focus $\left(3, 5\tfrac{1}{8}\right)$ , directrix $y = 4\tfrac{7}{8}$
The relationship is $a = \dfrac{1}{4p}$ . A large $a$ means a narrow parabola and a focus close to the vertex.

Completing the square on the conic form

A parabola in general form has only one squared variable:

$y^2 - 8y - 4x + 28 = 0$

Complete the square on the squared variable and make the other one the subject:
- $(y^2 - 8y + 16) = 4x - 28 + 16$
- $(y - 4)^2 = 4x - 12 = 4(x - 3)$
- Vertex $(3, 4)$ , $4p = 4$ so $p = 1$ , focus $(4, 4)$ , directrix $x = 2$ , opening right
Only one variable is squared — that is what distinguishes a parabola from the other three conics at a glance.

The reflective property in context

Every ray parallel to the axis reflects to the focus, and conversely a source at the focus produces a parallel beam.
In a design problem the useful chain is: given the width and depth of the dish, find $p$ , and place the receiver at the focus.
Worked through — a dish 4 m across and 1 m deep, vertex at the origin opening up:
- The rim is at $(2, 1)$ , so $x^2 = 4py$ gives $4 = 4p(1)$ , so $p = 1$
- The receiver goes 1 m above the vertex, on the axis.

Tips and Tricks

One squared variable means a parabola. Check that first — it saves attempting to force the equation into an ellipse or hyperbola form.

$4p$ , not $p$ , is the coefficient. From $y^2 = 12x$ the focal length is 3, not 12. Halving instead of quartering is the standard slip.

The parabola opens towards the focus. Sketch the vertex, mark the focus, and the direction of opening follows — it is never worth memorising four sign rules.

The directrix is a line, so give it as an equation ( $x = -3$ ), not as a point.

Use the latus rectum to sketch. The curve is $|4p|$ wide at the focus, which fixes two more points instantly and stops the sketch from being too narrow or too wide.

In a design context, "where does the receiver go?" always means the focus. State it as a distance from the vertex along the axis, in the context's own units.

Worked ExampleDesigning a solar concentrator

A parabolic solar trough has a cross-section 3.6 m wide and 0.9 m deep. Taking the vertex at the origin with the trough opening upwards, find the equation of the cross-section, the position of the collector pipe, and the width of the trough at the height of the pipe.

Step 1 — Set up the model

The trough opens upwards with vertex at the origin, so the form is

$x^2 = 4py$

with $p > 0$ and the focus at $(0, p)$ .

Step 2 — Use the dimensions to find a point on the curve

The trough is 3.6 m wide in total, so it extends 1.8 m each side of the axis, and it is 0.9 m deep.

Halving the width is the step that gets missed. The 3.6 m spans from $x = -1.8$ to $x = +1.8$ .

Step 3 — Substitute to find $p$

$(1.8)^2 = 4p(0.9)$

$3.24 = 3.6p$

$p = 0.9$

Step 4 — State the equation

$4p = 3.6, \qquad \text{so} \qquad x^2 = 3.6y$

Check the rim: $(1.8)^2 = 3.24$ and $3.6(0.9) = 3.24$ ✓

Step 5 — Locate the collector pipe

The pipe must sit at the focus, so that every ray arriving parallel to the axis is reflected onto it:

$\text{Focus} = (0,\ p) = (0,\ 0.9)$

Step 6 — Find the width at the height of the pipe

The width at the focus is the latus rectum, whose length is $|4p|$ :

$|4p| = 3.6 \text{ m}$

Confirming directly. At $y = 0.9$ :

$x^2 = 3.6(0.9) = 3.24 \quad \Rightarrow \quad x = \pm 1.8$

so the trough spans from $-1.8$ to $1.8$ , a width of 3.6 m ✓

Step 7 — Interpret and check the design

The design consequence. With the pipe at rim level:

Every ray entering parallel to the axis reaches the pipe ✓
The pipe is not shaded by the trough walls, since nothing rises above it
The supporting structure can rest on the rim rather than reaching out over the aperture

A design check. If the trough were made deeper without widening it — say 1.2 m deep, keeping 3.6 m wide — the rim would be at $(1.8, 1.2)$ , giving $3.24 = 4p(1.2)$ and $p = 0.675$ . The focus would then sit 0.675 m up, well below the 1.2 m rim, and the pipe would be recessed inside the trough — shading part of the aperture and losing efficiency. Deeper is not better for a parabolic collector, and the equation is what shows it.

The parabola

Covers AS91573

What defines a parabola

A parabola is the set of points that are the same distance from a fixed point (the focus) as from a fixed line (the directrix).
That definition is what makes parabolas useful. Every ray arriving parallel to the axis reflects off the curve and passes through the focus — which is why satellite dishes, headlight reflectors and solar concentrators are parabolic.
A parabola is unbounded: it opens out forever, unlike a circle or ellipse.

The two standard forms

Opening left or right (axis horizontal):

$(y - k)^2 = 4p(x - h)$

Opening up or down (axis vertical):

$(x - h)^2 = 4p(y - k)$

Each part:
- $(h, k)$ — the vertex, the turning point of the curve
- $p$ — the focal length: the distance from the vertex to the focus, and from the vertex to the directrix
- The squared variable tells you the axis. $y$ squared → horizontal axis; $x$ squared → vertical axis.
The sign of $p$ gives the direction of opening:

Form	$p > 0$	$p < 0$
$(y-k)^2 = 4p(x-h)$	opens right	opens left
$(x-h)^2 = 4p(y-k)$	opens up	opens down

Locating the focus and directrix

The parabola opens towards the focus and away from the directrix, and the vertex sits exactly halfway between them.

Form	Vertex	Focus	Directrix
$(y-k)^2 = 4p(x-h)$	$(h, k)$	$(h + p,\ k)$	$x = h - p$
$(x-h)^2 = 4p(y-k)$	$(h, k)$	$(h,\ k + p)$	$y = k - p$

Worked through — $y^2 = 12x$ $y^{2} = 12 x$ :
- Compare with $(y-k)^2 = 4p(x-h)$ : $h = 0$ , $k = 0$ , $4p = 12$ so $p = 3$
- Vertex $(0,0)$ , focus $(3, 0)$ , directrix $x = -3$ , opening right

The latus rectum

The latus rectum is the chord through the focus perpendicular to the axis.
Its length is $|4p|$ — exactly the coefficient in the standard form.
It is the fastest way to sketch the width of the curve. For $y^2 = 12x$ the parabola is 12 units wide at the focus, so it passes through $(3, 6)$ and $(3, -6)$ .

From the $y = ax^2 + bx + c$ you already know

The Level 2 form $y = ax^2 + bx + c$ $y = a x^{2} + b x + c$ is a parabola with a vertical axis. Completing the square converts it:
- $y = 2x^2 - 12x + 23$
- $y = 2(x^2 - 6x) + 23 = 2(x - 3)^2 - 18 + 23 = 2(x-3)^2 + 5$
- Rearranged into conic form: $(x - 3)^2 = \tfrac{1}{2}(y - 5)$
- So $4p = \tfrac{1}{2}$ , giving $p = \tfrac{1}{8}$ : vertex $(3, 5)$ , focus $\left(3, 5\tfrac{1}{8}\right)$ , directrix $y = 4\tfrac{7}{8}$
The relationship is $a = \dfrac{1}{4p}$ . A large $a$ means a narrow parabola and a focus close to the vertex.

Completing the square on the conic form

A parabola in general form has only one squared variable:

$y^2 - 8y - 4x + 28 = 0$

Complete the square on the squared variable and make the other one the subject:
- $(y^2 - 8y + 16) = 4x - 28 + 16$
- $(y - 4)^2 = 4x - 12 = 4(x - 3)$
- Vertex $(3, 4)$ , $4p = 4$ so $p = 1$ , focus $(4, 4)$ , directrix $x = 2$ , opening right
Only one variable is squared — that is what distinguishes a parabola from the other three conics at a glance.

The reflective property in context

Every ray parallel to the axis reflects to the focus, and conversely a source at the focus produces a parallel beam.
In a design problem the useful chain is: given the width and depth of the dish, find $p$ , and place the receiver at the focus.
Worked through — a dish 4 m across and 1 m deep, vertex at the origin opening up:
- The rim is at $(2, 1)$ , so $x^2 = 4py$ gives $4 = 4p(1)$ , so $p = 1$
- The receiver goes 1 m above the vertex, on the axis.

Tips and Tricks

One squared variable means a parabola. Check that first — it saves attempting to force the equation into an ellipse or hyperbola form.

$4p$ , not $p$ , is the coefficient. From $y^2 = 12x$ the focal length is 3, not 12. Halving instead of quartering is the standard slip.

The parabola opens towards the focus. Sketch the vertex, mark the focus, and the direction of opening follows — it is never worth memorising four sign rules.

The directrix is a line, so give it as an equation ( $x = -3$ ), not as a point.

Use the latus rectum to sketch. The curve is $|4p|$ wide at the focus, which fixes two more points instantly and stops the sketch from being too narrow or too wide.

In a design context, "where does the receiver go?" always means the focus. State it as a distance from the vertex along the axis, in the context's own units.

Worked ExampleDesigning a solar concentrator

Step 1 — Set up the model

The trough opens upwards with vertex at the origin, so the form is

$x^2 = 4py$

with $p > 0$ and the focus at $(0, p)$ .

Step 2 — Use the dimensions to find a point on the curve

The trough is 3.6 m wide in total, so it extends 1.8 m each side of the axis, and it is 0.9 m deep.

Halving the width is the step that gets missed. The 3.6 m spans from $x = -1.8$ to $x = +1.8$ .

Step 3 — Substitute to find $p$

$(1.8)^2 = 4p(0.9)$

$3.24 = 3.6p$

$p = 0.9$

Step 4 — State the equation

$4p = 3.6, \qquad \text{so} \qquad x^2 = 3.6y$

Check the rim: $(1.8)^2 = 3.24$ and $3.6(0.9) = 3.24$ ✓

Step 5 — Locate the collector pipe

The pipe must sit at the focus, so that every ray arriving parallel to the axis is reflected onto it:

$\text{Focus} = (0,\ p) = (0,\ 0.9)$

Step 6 — Find the width at the height of the pipe

The width at the focus is the latus rectum, whose length is $|4p|$ :

$|4p| = 3.6 \text{ m}$

Confirming directly. At $y = 0.9$ :

$x^2 = 3.6(0.9) = 3.24 \quad \Rightarrow \quad x = \pm 1.8$

so the trough spans from $-1.8$ to $1.8$ , a width of 3.6 m ✓

Step 7 — Interpret and check the design

The design consequence. With the pipe at rim level:

Every ray entering parallel to the axis reaches the pipe ✓
The pipe is not shaded by the trough walls, since nothing rises above it
The supporting structure can rest on the rim rather than reaching out over the aperture

The parabola

What defines a parabola

The two standard forms

Locating the focus and directrix

The latus rectum

From the y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c you already know

Completing the square on the conic form

The reflective property in context

Step 1 — Set up the model

Step 2 — Use the dimensions to find a point on the curve

Step 3 — Substitute to find ppp

Step 4 — State the equation

Step 5 — Locate the collector pipe

Step 6 — Find the width at the height of the pipe

Step 7 — Interpret and check the design

The parabola

What defines a parabola

The two standard forms

Locating the focus and directrix

The latus rectum

From the y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c you already know

Completing the square on the conic form

The reflective property in context

Step 1 — Set up the model

Step 2 — Use the dimensions to find a point on the curve

Step 3 — Substitute to find ppp

Step 4 — State the equation

Step 5 — Locate the collector pipe

Step 6 — Find the width at the height of the pipe

Step 7 — Interpret and check the design

From the $y = ax^2 + bx + c$ you already know

Step 3 — Substitute to find $p$

From the $y = ax^2 + bx + c$ you already know

Step 3 — Substitute to find $p$