The Parabola (Edexcel IAL Further Maths)

Equation of a Parabola

Spec point — 4.1

The Equation of a Parabola

What is a parabola?

• A parabola is a U-shaped quadratic curve with a line of symmetry

  • $y=x^{2}$* *is a parabola

    • The line of symmetry is the $y$-axis

  • $x=y^{2}$* *is a parabola

    • The line of symmetry is the $x$-axis

• A parabola is part of a family of curves called the conics (or conic sections)

  • Conics are parabolae, hyperbolae and ellipses

What is the general equation of a parabola?

The general parabola

• The general equation for a parabola is

  • $y^{2}=4ax$ in Cartesian form

  • $x=at^{2},y=2at$ in Parametric form

    • where $a$ is a positive constant

• The $x$-axis is the line of symmetry, as shown

• The general equation can be rearranged

  • $y=\pm\sqrt{4ax}$

    • $y=\sqrt{4ax}$ is the branch above the x-axis
    • $y=-\sqrt{4ax}$ is the branch below the x-axis

Exam Hint You are given the Cartesian and parametric equations of a parabola in the Formulae Booklet.

Worked Example A parabola has the parametric equations $x=at^{2}$* *and $y=2at$ where $a>0$.

Show that its Cartesian equation is $y^{2}=4ax$.

To find the Cartesian equation, eliminate $t$ from the parametric equations It is easier to make $t$ the subject of $y=2at$

$t=\frac{y}{2a}$

Substitute this into the equation for $x$ and simplify

$\begin{gathered} x = a ( \frac{y}{2 a} )^{2} \\ x = \frac{a y^{2}}{4 a^{2}} \\ x = \frac{y^{2}}{4 a} \end{gathered}$

Make $y^{2}$ the subject

$y^{2}=4ax$

Focus & Directrix of a Parabola

Spec point — 4.3

The Focus & Directrix of a Parabola

What are the focus and directrix of a parabola?

• The focus of the parabola $y^{2}=4ax$ is the point $( a , 0 )$ on the $x$-axis

• The directrix is the vertical line$x=-a$

• For example, the parabola $y^{2}=12x$ where $a=3$ has

  • a focus at $( 3 , 0 )$
  • the directrix $x=-3$

What is the focus-directrix property?

• The focus-directrix property says that:

  • “Points on a parabola are the same distance from the focus as they are horizontally from the directrix”

• In other words:

  • If S is the focus of a parabola

  • and P is any point on the parabola

  • and X is the point on the directrix horizontally from P

  • Then the distance PS equals PX

    • PS = PX

• This property is an example of a locus of points

Exam Hint You are given the focus and directrix of a parabola in the Formulae Booklet, but not the focus-directrix property.

Worked Example The diagram shows the point* *$S ( a , 0 )$, the vertical line $x=-a$ and a general point $P ( x , y )$ that can move in the plane.

If $P$ is restricted to always be the same distance from $S$ as it is horizontally from the vertical line, use coordinate geometry to prove that it must lie on the parabola $y^{2}=4ax$.

It helps to add the point X on to the diagram, on the line $x=-a$ horizontally from P The question says that PS = PX *The distance PS can be found using Pythagoras' theorem *

$P S^{2} = ( x - a )^{2} + y^{2}$

The distance PX can be found from the diagram

$PX=x+a$

Square PS= PX and substitute in the results above

$\begin{gathered} P S^{2} & = P X^{2} \\ ( x - a )^{2} + y^{2} & = ( x + a )^{2} \end{gathered}$

Expand and simplify both sides

$\begin{gathered} x^{2} - 2 a x + a^{2} + y^{2} & = x^{2} + 2 a x + a^{2} \\ - 2 a x + y^{2} & = 2 a x \\ y^{2} & = 4 a x \end{gathered}$

$P ( x , y )$ must lie on this curve

$y^{2}=4ax$