Coordinate systems (Further Pure 1)

Explanation

• Four standard conics carry standard parametrisations.
• The parabola $y^2=4ax$ is written as $\left(at^2,\,2at\right)$; the rectangular hyperbola $xy=c^2$ as $\left(ct,\,\dfrac{c}{t}\right)$; the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ as $\left(a\cos t,\,b\sin t\right)$; and the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ as $\left(a\sec t,\,b\tan t\right)$ or $\left(a\cosh t,\,b\sinh t\right)$.
• To convert a parametrisation to Cartesian form, eliminate the parameter using the identity that matches it: $\cos^2t+\sin^2t=1$, $1+\tan^2t=\sec^2t$ or $\cosh^2t-\sinh^2t=1$.
• Translating a conic replaces $x$ by $x-p$ and $y$ by $y-q$ throughout, moving the centre or vertex to $(p,q)$ without changing the shape.
• Note that $\left(a\cosh t,\,b\sinh t\right)$ traces only the branch with $x\ge a$, because $\cosh t\ge1$, whereas the secant-tangent form covers both branches.

Explanation

• Every conic can be defined by a focus $S$, a directrix $\ell$ and an eccentricity $e$, as the locus of points $P$ with $SP=e\times(\text{distance from }P\text{ to }\ell)$. For the parabola $y^2=4ax$, $e=1$, the focus is $(a,0)$ and the directrix is $x=-a$.
• For the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with $a>b$, $e<1$ and $b^2=a^2\left(1-e^2\right)$, the foci are $(\pm ae,0)$ and the directrices are $x=\pm\dfrac{a}{e}$.
• For the hyperbola $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, $e>1$ and $b^2=a^2\left(e^2-1\right)$, with the same expressions for the foci and directrices.
• The focal distance of a point on an ellipse is $SP=a-ex$, so $SP+S'P=2a$; on a hyperbola it is $SP=ex-a$ on the branch nearer the focus.
• Learn which of $1-e^2$ and $e^2-1$ belongs to which curve; that single sign is the most common source of lost marks.

Explanation

• Tangents to a parametrised conic are found by implicit or parametric differentiation and then written in a standard form worth memorising. For $y^2=4ax$ at $\left(at^2,2at\right)$ the tangent is $ty=x+at^2$ and the normal is $y+tx=2at+at^3$.
• For $xy=c^2$ at $\left(ct,\dfrac{c}{t}\right)$ the tangent is $x+t^2y=2ct$ and the normal is $t^3x-ty=c\left(t^4-1\right)$.
• For the ellipse at $\left(a\cos\theta,b\sin\theta\right)$ the tangent is $\dfrac{x\cos\theta}{a}+\dfrac{y\sin\theta}{b}=1$, and for the hyperbola at $\left(a\sec\theta,b\tan\theta\right)$ it is $\dfrac{x\sec\theta}{a}-\dfrac{y\tan\theta}{b}=1$.
• You are also expected to know the tangency conditions: $y=mx+c$ touches $y^2=4ax$ when $c=\dfrac{a}{m}$, touches $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ when $c^2=a^2m^2+b^2$, and touches $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$ when $c^2=a^2m^2-b^2$; for the rectangular hyperbola $xy=c^2$ the line $y=mx+c_1$ touches it when $c_1^2+4mc^2=0$, which can only happen when $m<0$.
• All four are expected to be known, and all four come from setting the discriminant of the resulting quadratic to zero, which is also how you find tangents through an external point.

Explanation

• A locus question asks for the equation of the curve traced by a moving point, and the reliable method is always the same three steps. First, write the coordinates of the moving point in terms of a parameter, usually $t$ or $\theta$ from the parametrisation of the underlying conic.
• Second, set $x$ and $y$ equal to those two expressions. Third, eliminate the parameter between them to leave a relation in $x$ and $y$ alone.
• Where a geometric condition is imposed on a chord, first turn it into an algebraic relation between the two parameters, such as $pq=-1$ for a focal chord of $y^2=4ax$ or $pq=-4$ for a chord subtending a right angle at the vertex.
• Symmetric functions help: if only $p+q$ and $pq$ appear, use $p^2+q^2=(p+q)^2-2pq$.
• Finally, state any restriction on the range of the locus that the geometry forces, since a bare equation may describe more of a curve than is actually traced.