Properties of Parabolas (Edexcel A Level Further Maths: Further Pure 1): Revision Note

You are given the Cartesian equation of a parabola in the formulae booklet.

You are given the parametric equations of a parabola in the formulae booklet.

You are given the eccentricity, focus and directrix of a parabola in the formulae booklet.

A parabola has the equation $y^2=20x$.

Calculate

(a) the coordinates of the focus,

(b) the equation of the directrix.

Answer:

(a)

Find $a$ by comparing to the general equation $y^2=4ax$

$a=5$

Substitute into $\left(a,0\right)$

The focus has coordinates $\left(5,0\right)$

(b)

Substitute $a=5$ into the equation of a directrix, $x=-a$

The directrix has the equation $x=-5$

You are not given the focus-directrix property in the exam (you must learn it).

A parabola with focus $F$ at $\left(a,0\right)$and directrix $x=-a$ is shown below.

The point $P$ on the parabola has coordinates $\left(x,y\right)$ and the point $D$ is on the directrix, at the same height as $P$.

Using only the focus-directrix property, derive the Cartesian equation of a parabola, $y^2=4ax$.

Answer:

Use the focus-directrix property on $P$, $F$ and $D$ (draw on the lines $PF$ and $PD$)

$\frac{PF}{PD}=1$

It helps to draw the lengths $x$ and $y$ from $P\left(x,y\right)$ on the diagram

Create a right-angled triangle whose hypotenuse is $PF$ with base $\left(x-a\right)$ and height $y$

Use Pythagoras' theorem to find $P{F}^2$

$P{F}^2={\left(x-a\right)}^2+y^2$

Find the length $PD$ from $x$ to the directrix

$PD=x-\left(-a\right)=x+a$

Rearrange $\frac{PF}{PD}=1$ to make $P{F}^2$ the subject

$P{F}^2=P{D}^2$

Substitute in expressions for $P{F}^2$ and $P{D}^2$ from above

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

Expand and cancel

$\begin{array}{rcl}{x}^2-2ax+a^2+y^2& =& x^2+2ax+a^2\\ \cancel{x^2}-2ax+\bcancel{a^2}+y^2& =& \cancel{x^2}+2ax+\bcancel{a^2}\\ -2ax+y^2& =& 2ax\end{array}$

Add $2ax$ to both sides

The Cartesian equation is $y^2=4ax$