## The Parabola

Parabola definition: locus of $P$ such that $PS=PL$.

### Cartesian Equations: $(x-a)^{2}+y^{2}=(x+a)^{2}$ $x^{2}-2ax+a^{2}+y^{2} = x^{2}+2ax+a^{2}$ $y^{2} = 4ax$ $2y\dfrac{\textnormal{d}y}{\textnormal{d}x}=4a\ \implies\ \dfrac{\textnormal{d}y}{\textnormal{d}x}=\dfrac{2a}{y}$

### Parametric Equation: $x=at^{2},\ y=2at$ $\dfrac{\textnormal{d}x}{\textnormal{d}t}=2at,\ \dfrac{\textnormal{d}y}{\textnormal{d}t}=2a\ \implies\ \dfrac{\textnormal{d}y}{\textnormal{d}x}=\dfrac{1}{t}$

tangent: $y-2at=\dfrac{1}{t}(x-at^{2})\ \implies ty=x+at^{2}$

normal: $y-2at=-t(x-at^{2})\ \implies y+tx=2at+at^{3}$

## t-Formulae and parameterisation of the circle

t-Formulae are used in integration to tackle rational expressions of tigonometric functions.  After a spell in the cold, when they were not included in some A level specifications, they are now back in sixth form lessons.

It all starts with the subsitution, $t=\tan \frac{\theta}{2}$

from which the following functions can be derived, $\tan \theta=\dfrac{2t}{1-t^{2}}$ $\cos \theta =\dfrac{1-t^{2}}{1+t^{2}}$ $\sin \theta=\dfrac{2t}{1+t^{2}}$.

These derivations can be made using compound trigonmetry fomulae.  Alternatively, there is an engaging co-ordinate geometry derivation which has the merits of doubling up as an algebraic parametrisation of the circle. Euclid tells us that the angle subtended by the chord $PP'$ at the centre is twice the angle subtended at the circumference.  The $X$-axis providing a line of symmetry, gives the relationship between the angles $\frac{\theta}{2}$ and $\theta$ at $A$ and $O$ respectively.

Defining $t=\tan \theta$ and creating a line, $y=t(x+1)$, through $A$ with gradient $t$ gives intersections with the $Y$-axis and the circle at $R$ and $P$ respectively.

The intersection point $P$ can then be found by solving the simulataneous equations: $y=t(x+1)$, and, $x^2+y^2=1$.

Substituting for $y$ leads to the quadratic, $(t^2+1)x^2+2t^2x+t^2-1=0$,

which admits an easy factorisation once one acknowledges that it must have one root of $x=-1$. $(x+1)(x-\frac{1-t^2}{1+t^2})=0$,

giving the other root, and $x$ value for $P$ as $\frac{1-t^2}{1+t^2}$.

Solving for $y$ gives the co-ordinates of $P$.  When viewed as two altenative parameterisations of the unit circle, the derivation of the $t$ formulae is complete. $\left(\dfrac{1-t^{2}}{1+t^{2}},\dfrac{2t}{1+t^{2}}\right)=(\cos \theta,\sin \theta )$.