The parabola from focus and directrix

What you will be able to do

• say what a parabola is and write its equation from the focus and the directrix;
• read the vertex, focus, directrix, axis and latus rectum from any of the four standard forms;
• use the focal distance and the parametric form $(at^2, 2at)$;
• recognise a parabola when its vertex is not at the origin.

A ball and a dish

A ball thrown through the air follows a parabola. A satellite dish has the shape of a parabola turned round its axis, and the signal collects at one special point in front of the dish. That point is the focus. A parabola has a short definition that explains all this: it is the set of points that are the same distance from a fixed point and from a fixed line.

You have seen parabolas as graphs of $y = ax^2 + bx + c$ since Class 9. Here we study them as geometry, and we give the special point a name.

The definition

Fix a point $F$, the focus, and a line $l$ that does not pass through $F$, the directrix. The parabola is the set of all points $P$ for which

where $PM$ is the perpendicular distance from $P$ to $l$. We say that the eccentricity is $e = 1$. The eccentricity is the ratio $PF : PM$. In the next course the ellipse has $e < 1$ and the hyperbola has $e > 1$, so the parabola is the middle case.

Some words. The line through the focus perpendicular to the directrix is the axis of the parabola. The point of the parabola halfway between the focus and the directrix on the axis is the vertex.

The equation $y^2 = 4ax$

Put the vertex at the origin, the focus at $F(a, 0)$ and the directrix as the line $x = -a$, with $a > 0$. A point $P(x, y)$ is on the parabola when

Square both sides: $x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2$. The terms $x^2$ and $a^2$ cancel, and we are left with

The focal distance of a point $P(x_1, y_1)$ on this parabola is $PF = PM = x_1 + a$.

Four standard forms

Turn the parabola round and you get three more.

• $y^2 = 4ax$: opens to the right. Focus $(a, 0)$, directrix $x = -a$.
• $y^2 = -4ax$: opens to the 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 $y = a$.

In all four the vertex is $(0, 0)$ and $a > 0$. The axis is the $x$-axis for the first two and the $y$-axis for the last two. The sign in front of $4a$ tells you the direction. A plus sign opens towards the positive axis and a minus sign opens towards the negative axis.

The latus rectum

The latus rectum is the chord through the focus perpendicular to the axis. For $y^2 = 4ax$ put $x = a$: $y^2 = 4a^2$, so $y = \pm 2a$. The ends are $(a, 2a)$ and $(a, -2a)$ and the length is $4a$. It is the width of the parabola at the focus, and it measures how open the parabola is.

A worked example

Take $y^2 = 12x$. Then $4a = 12$, so $a = 3$.

1. Focus and directrix. The focus is $(3, 0)$ and the directrix is $x = -3$.
2. Latus rectum. It has length $4a = 12$, with ends $(3, 6)$ and $(3, -6)$. Check: $6^2 = 36 = 12 \times 3$.
3. A point. The point $(12, 12)$ is on the parabola. Its distance from the focus is $\sqrt{9^2 + 12^2} = 15$. Its distance from the directrix is $12 + 3 = 15$. They agree.

Now take $x^2 = -8y$. Here $4a = 8$, so $a = 2$. The parabola opens downward. The focus is $(0, -2)$ and the directrix is $y = 2$.

A parabola from its focus and directrix

Let the focus be $(2, 1)$ and the directrix the line $y = -3$. Write $PF^2 = PM^2$ for $P(x, y)$:

Expand the right side and cancel $y^2$: $(x - 2)^2 + 1 - 2y = 6y + 9$, so $(x - 2)^2 = 8y + 8 = 8(y + 1)$. The vertex is halfway between the focus and the foot $(2, -3)$ of the perpendicular on the directrix, so it is $(2, -1)$. Here $a = 2$. In general

is a parabola with the vertex $(h, k)$, the axis $y = k$, the focus $(h + a, k)$ and the directrix $x = h - a$. The other three forms shift in the same way, for instance $(x - h)^2 = 4a(y - k)$.

The general quadratic and the parametric form

Any equation with one variable squared and the other to the first power, such as $y^2 + 4y - 8x + 12 = 0$, is a parabola. Complete the square: $(y + 2)^2 = 8x - 8 = 8(x - 1)$. The graph of $y = px^2 + qx + r$ from algebra is also a parabola with a vertical axis. For example, $y = \dfrac{x^2}{8}$ is $x^2 = 8y$, with $a = 2$.

Every point of $y^2 = 4ax$ can be written as

Check: $y^2 = 4a^2t^2 = 4a \times at^2 = 4ax$. The value $t = 0$ is the vertex and $t = \pm 1$ are the ends of the latus rectum. The focal distance of this point is $x + a = a(t^2 + 1)$.

A point $(x_1, y_1)$ is outside the parabola (on the side away from the focus) when $y_1^2 - 4ax_1 > 0$, on it when $= 0$ and inside when $< 0$. The point $(1, 1)$ gives $1 - 4 = -3$ for $y^2 = 4x$, so it is inside.

Practise with new numbers

Try it in Python

The program takes a parabola $y^2 = Kx$ and finds the focus, the directrix and the latus rectum. It then checks the definition at several points.

Common mistakes

• Reading $a$ as the coefficient. In $y^2 = 12x$ the number $12$ is $4a$, so $a = 3$.
• Putting the focus on the wrong axis. For $y^2 = 4ax$ the focus is on the $x$-axis. For $x^2 = 4ay$ it is on the $y$-axis.
• Forgetting the sign of the directrix. For $y^2 = 4ax$ the directrix is $x = -a$, on the opposite side from the focus.
• Using $x$ for the focal distance when the parabola is shifted. Measure from the directrix of the shifted parabola.

Remember these

• A parabola: $PF = PM$, with $e = 1$.
• $y^2 = 4ax$: focus $(a, 0)$, directrix $x = -a$, latus rectum $4a$, focal distance $x_1 + a$.
• Four forms: $y^2 = \pm 4ax$ and $x^2 = \pm 4ay$.
• $(y - k)^2 = 4a(x - h)$ has the vertex $(h, k)$.
• The parametric form is $(at^2, 2at)$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the focus and the directrix of $y^2 = 16x$.

2. Find the focal distance of the point at $t = 3$ on $y^2 = 4x$.

3. Find the focus, the directrix and the latus rectum of $x^2 = -12y$.

4. Find the equation of the parabola with the focus $(0, 2)$ and the directrix $y = -2$.

5. The parabola $y^2 = 4ax$ passes through $(3, 6)$. Find $a$.

References

• NCERT, Mathematics for Class XI, Chapter 10 (Conic Sections), the section on the parabola.
• OpenStax, Precalculus 2e, Chapter 12 (Analytic Geometry), the section on the parabola.