The parabolaEdexcel International A Level Further Maths: Revision notes

The parabola $y^2=4ax$, with $a>0$, has its vertex at the origin and the $x$-axis as its axis of symmetry, opening to the right. It has

• focus $S(a,0)$,
• directrix $l$: $x=-a$.

For $y^2=12x$, $4a=12$ so $a=3$, the focus is $(3,0)$ and the directrix is $x=-3$. The curve is symmetric about the $x$-axis: for every $x>0$ there are two values $y=\pm\sqrt{4ax}$.

Every point on $y^2=4ax$ can be written $(at^2,2at)$ for a parameter $t$: $x=at^2,\qquad y=2at.$ Check: $y^2=4a^2t^2=4a(at^2)=4ax$. The vertex is $t=0$, and $t$ and $-t$ give points that are reflections in the $x$-axis. For $y^2=8x$, $a=2$, the point with $t=-3$ is $(2\cdot9,\,4\cdot(-3))=(18,-12)$. To find $t$ from a point, use the $y$-coordinate $y=2at$, since $x=at^2$ only gives $t$ up to sign.

A parabola is the locus of points equidistant from a fixed point (the focus) and a fixed line (the directrix). For $P$ on $y^2=4ax$ with focus $S$: $SP=\text{distance from }P\text{ to the directrix}=x_P+a.$ Proof of the locus: for $Q(x,y)$, $\sqrt{(x-a)^2+y^2}=x+a$. Squaring gives $x^2-2ax+a^2+y^2=x^2+2ax+a^2$, so $y^2=4ax$. For $P(at^2,2at)$: $SP=at^2+a=a(t^2+1)$, which agrees with $\sqrt{(at^2-a)^2+(2at)^2}$.

$C$: $y^2=20x$, so $a=5$, $S=(5,0)$, $l$: $x=-5$. A point $P$ in the first quadrant has $SP=30$. Then $x+5=30$, so $x=25$ and $y^2=500$, giving $P=(25,10\sqrt5)$. The triangle $OSP$ has base $OS=5$ and height $10\sqrt5$, so its area is $\frac12(5)(10\sqrt5)=25\sqrt5$. To test whether $T(10,15)$ is on $C$: $TS=\sqrt{250}\approx15.8$ but the distance to $l$ is $15$, so $T$ is not on $C$.