The parabola in Year 3: the class test

The parabola test almost always falls between the end of November and the middle of December, right before the end-of-term grading meeting. It is the first real coordinate geometry test of the last three years of high school, and anyone who arrives unprepared pays twice: on the December school report and again in January, because the method for tangents that you learn here is identical for the circle, the ellipse and the hyperbola. Whoever understands it in December saves three months.

The misunderstanding I see most often is believing that the parabola is “quadratic equations with a drawing”. It is not. In Year 2 you looked for the values of $x$ that make a trinomial zero; here you look for points, that is, pairs of coordinates, and almost every exercise in the test comes down to a single idea repeated: put a line and the parabola into a system and look at the discriminant.

Here you will find the formulas you really need, the method for tangents in the three versions that come up in the test, three worked exercises and the five mistakes that cost the most marks.

The formulas you need (and the ones you can derive)

The equation of a parabola with axis parallel to the $y$ axis is

y = a x^{2} + b x + c (a \neq = 0)

The formulas to have on automatic are three, not ten:

V (- \frac{b}{2 a}, - \frac{Δ}{4 a}), axis: x = - \frac{b}{2 a}, Δ = b^{2} - 4 a c

The $x$ coordinate of the vertex is the one that matters: you can always get the ordinate by substituting that $x$ into the equation, without remembering $- \frac{Δ}{4 a}$ . If you are in doubt during the test, substitute: it costs ten seconds and it is impossible to get wrong.

The sign of $a$ gives the concavity: upwards if $a > 0$ , downwards if $a < 0$ . The term $c$ is the ordinate of the point where the curve meets the $y$ axis, because for $x = 0$ only $c$ is left.

The focus and the directrix are needed less often, but in many classes they appear in a theory question:

F (- \frac{b}{2 a}, \frac{1 - Δ}{4 a}), directrix: y = - \frac{1 + Δ}{4 a}

Finally, when the axis is parallel to the $x$ axis the equation is $x = a y^{2} + b y + c$ : all the formulas hold with $x$ and $y$ swapped, nothing else to learn. This is the classic exercise where whoever memorised the formulas freezes and whoever understood the symmetry solves it in a minute.

Finding a parabola: as many conditions as unknowns

The equation has three parameters, $a$ , $b$ , $c$ : you need three independent conditions to find it, and each one becomes an equation of the system. The typical conditions in the test:

Condition in the question	Equation you write
passes through $P (x_{0}, y_{0})$	$y_{0} = a x_{0}^{2} + b x_{0} + c$
has vertex at $V (x_{V}, y_{V})$	counts for two: $- \frac{b}{2 a} = x_{V}$ and passing through $V$
has axis $x = k$	$- \frac{b}{2 a} = k$
meets the $x$ axis at $x_{1}$ and $x_{2}$	it is best to write it as $y = a (x - x_{1}) (x - x_{2})$
is tangent to a given line	$Δ = 0$ of the line-parabola system

Two warnings that are worth marks. The vertex counts for two conditions, not one: whoever treats it as a simple point to pass through ends up with two equations and three unknowns. And when you know the two intersections with the $x$ axis, the form $y = a (x - x_{1}) (x - x_{2})$ reduces the exercise to a single unknown: it is the most useful shortcut in the chapter.

Tangents: the whole test is one idea

Put the line and the parabola into a system, get a quadratic equation in $x$ , and look at its discriminant:

$Δ > 0$ → the line is a secant, two points in common;
$Δ = 0$ → the line is a tangent, one (double) point;
$Δ < 0$ → the line is external, no points.

This discriminant is not the $Δ$ of the parabola. They are two different numbers that go by the same name: the $Δ$ of the parabola tells you how many times the curve cuts the $x$ axis, this one tells you how many times the line you are studying meets it. Mixing them up is mistake number one in the December test. It is worth calling them $Δ_{par}$ and $Δ_{sist}$ on the page until it becomes automatic.

The three versions that come up in the test:

Tangents from an external point $P$ . Write the pencil (the family) of lines through $P$ , that is $y - y_{P} = m (x - x_{P})$ , put it into a system and impose $Δ_{sist} = 0$ : you get a quadratic equation in $m$ and therefore two tangents. A note that teachers appreciate: the pencil written this way excludes the vertical line $x = x_{P}$ , and you should say that the exclusion loses no solutions, because a vertical line meets a vertical-axis parabola at a single point but is not tangent, it crosses it.
Tangent parallel to a given line. The slope $m$ is known, the unknown is $q$ : write $y = m x + q$ and impose $Δ_{sist} = 0$ . One equation, one solution.
Tangent at a point $P$ of the parabola. Here the doubling formula works, which avoids the system:

\frac{y + y _{0}}{2} = a x x_{0} + b \frac{x + x _{0}}{2} + c

You get it from the equation of the parabola by replacing $y$ with $\frac{y + y _{0}}{2}$ , $x^{2}$ with $x x_{0}$ and $x$ with $\frac{x + x _{0}}{2}$ . It works only if $P$ lies on the parabola: using it for an external point gives a line that has nothing to do with the problem.

Pencils of parabolas

A pencil (a one-parameter family) is written by combining two parabolas (or a parabola and a line) with a parameter:

y = a x^{2} + b x + c + k (a^{'} x^{2} + b^{'} x + c^{'})

The questions in the test are always the same three: find the base points (the intersections common to all the curves of the pencil, which you get by putting the two generating curves into a system), find the curve of the pencil that satisfies one more condition (passing through a point, vertex on a line, tangency), and recognise the degenerate curves, that is, the values of $k$ for which the coefficient of $x^{2}$ vanishes and the parabola becomes a line.

This last point is the one people forget: if the question says “for which values of $k$ do you get a parabola”, the answer begins with $a + k a^{'} \neq = 0$ .

Three worked exercises

1. Parabola through three points

Find the parabola through $A (0, 3)$ , $B (1, 0)$ and $C (3, 0)$ , and determine the vertex.

$B$ and $C$ are on the $x$ axis: they are the two intersections, so the factorised form is the best choice

y = a (x - 1) (x - 3)

One unknown is left. I impose passing through $A (0, 3)$ :

3 = a (0 - 1) (0 - 3) = 3 a \Rightarrow a = 1

Expanding: $y = (x - 1) (x - 3) = x^{2} - 4 x + 3$ .

The vertex has abscissa $x_{V} = - \frac{- 4}{2} = 2$ , and substituting, $y_{V} = 4 - 8 + 3 = - 1$ . So $V (2, - 1)$ , concave upwards because $a = 1 > 0$ .

With the system in $a$ , $b$ , $c$ the same exercise needed three equations: the factorised form needed one.

2. Tangents from an external point

Find the lines tangent to the parabola $y = x^{2}$ drawn from $P (0, - 1)$ .

The point has a negative ordinate and the parabola has its vertex at the origin: $P$ is external. I write the pencil of lines through $P$ :

y = m x - 1

I put it into a system with the parabola and bring everything to the left-hand side:

x^{2} = m x - 1 \Rightarrow x^{2} - m x + 1 = 0

I impose tangency:

Δ_{sist} = m^{2} - 4 = 0 \Rightarrow m = \pm 2

The two tangents are therefore $y = 2 x - 1$ and $y = - 2 x - 1$ .

The points of tangency are found by solving the equation with the zero $Δ$ , that is $x = \frac{m}{2}$ : for $m = 2$ we get $x = 1$ , for $m = - 2$ we get $x = - 1$ . The points are $(1, 1)$ and $(- 1, 1)$ , and they are symmetric about the axis of the parabola. This is a consistency check that is always worth doing.

3. Tangent parallel to a given line

Find the line tangent to $y = x^{2} - 4 x + 3$ and parallel to $y = 2 x + 5$ .

Parallel means the same slope: $m = 2$ . The only unknown is $q$ .

y = 2 x + q

System with the parabola:

x^{2} - 4 x + 3 = 2 x + q \Rightarrow x^{2} - 6 x + (3 - q) = 0

Tangency condition:

Δ_{sist} = 36 - 4 (3 - q) = 24 + 4 q = 0 \Rightarrow q = - 6

The tangent is $y = 2 x - 6$ . The point of tangency has abscissa $x = \frac{6}{2} = 3$ and ordinate $y = 2 \cdot 3 - 6 = 0$ ; on the parabola: $9 - 12 + 3 = 0$ . It checks out.

The five mistakes that cost the most marks

1. Mixing up the two discriminants. The $Δ$ of the parabola concerns the intersections with the $x$ axis; the $Δ$ of the system concerns the intersections with the line you are studying. They are different and serve different purposes.

2. Forgetting the minus sign in the vertex. The abscissa is $- \frac{b}{2 a}$ . With a negative $b$ the double sign is confusing: in $y = x^{2} - 4 x + 3$ the abscissa is $+ 2$ , not $- 2$ .

3. Treating the vertex as a single condition. It counts for two. This is why many systems in the test “do not work out”.

4. Using the doubling formula with an external point. It holds only for points that lie on the parabola. For an external point you need the pencil of lines and $Δ = 0$ .

5. Not discussing the case $a = 0$ in parameter problems. If the coefficient of $x^{2}$ depends on $k$ , there is a value of $k$ that makes it vanish: there the curve is no longer a parabola but a line, and you must say so explicitly. It is half a mark that almost nobody gets.

The two weeks before the test

Rewrite the three vertex formulas on a blank sheet every day, and next to them their version with the axis parallel to the $x$ axis. Two minutes.
Do ten tangency exercises divided by type: five from an external point, three parallel to a given line, two with the doubling formula. In blocks, not mixed.
Do five “find the parabola” exercises writing only the system, without solving it. The mistake, in almost all cases, is in the set-up, not in the calculation, and this exercise trains it in a quarter of the time.
In the last two days, one timed practice paper, marked by sorting each mistake into three categories: wrong set-up, wrong calculation, forgotten condition. They are three different problems and three different remedies.

If what freezes is not the preparation but your head in the first minutes, I have written about it in how to manage test anxiety in maths and physics.

How we work on this in lessons

On the parabola I insist on one thing only: set up without solving. We look at the question, write the system, and move on to the next exercise without doing the calculations. Ten exercises set up in half an hour teach more than three solved in full, because the December test is almost always lost in translating the question, not in the algebra.

The other thing I work on is the link forward: the $Δ = 0$ method you learn here is the same one you will use on the circle in January and on the ellipse and hyperbola in spring. Whoever absorbs it now arrives in March with half the programme already in hand. Lessons are online and one-to-one: the costs (single lesson or 10-hour package) are on the pricing page, useful for comparison if it is better to cover the whole block on conic sections rather than a single test. The circle, with the same method, is covered for a different school track in the circle in the Cartesian plane; where pencils come from, that is from lines, is the starting point of pencils of lines: proper and improper.

The fastest way to see where you stand is to try the practice test on the parabola: fifteen questions in fifty minutes, with a breakdown of your weak points by topic at the end. The correct answer to each question is free and needs no registration; the step-by-step guided solution unlocks with your first lesson. If after the test you see that the tangents hold up but the pencils do not (or the other way round), write to me for a free introductory call: we look at the programme of your test and I tell you honestly how many lessons you need. The full picture of how I set up the course is on the maths tutoring page.

The parabola: tangents, pencils and parameter problems