Parabola and rectangular hyperbolaEdexcel A-Level Further Maths: Revision notes

The standard parabola with its vertex at the origin and the $x$-axis as its axis of symmetry has Cartesian equation $y^2=4ax,\quad a>0.$ Its focus is $S(a,0)$ and its directrix is the line $x=-a$. Every point can be written in parametric form $x=at^2,\quad y=2at,$ where the parameter $t$ labels the point. Eliminating $t$: $y^2=4a^2t^2=4a(at^2)=4ax$. Reading off $a$ from an equation is the first step in any question: for $y^2=20x$, $4a=20$ so $a=5$, the focus is $(5,0)$ and the directrix is $x=-5$. The point with parameter $t=2$ is $(20,20)$.

A parabola is the locus of points $P$ whose distance from the focus equals the perpendicular distance to the directrix: $PS=PM$, where $M$ is the foot of the perpendicular from $P$ to the directrix. For $P=(x,y)$ on $y^2=4ax$: $\sqrt{(x-a)^2+y^2}=x+a.$ Squaring gives $(x-a)^2+y^2=(x+a)^2$, so $y^2=4ax$, which proves the property. A useful consequence is that the distance from any point on the parabola to the focus is $x+a$. For $y^2=20x$ and a point with $x=20$, $PS=20+5=25$, which saves calculating $y$.

A rectangular hyperbola has the coordinate axes as its asymptotes. Its Cartesian equation is $xy=c^2,$ with parametric form $x=ct,\quad y=\frac{c}{t},\quad t\neq0.$ The general point is $\left(ct,\frac ct\right)$, and the product of the coordinates is $c^2$. The curve has two branches: $t>0$ is in the first quadrant and $t<0$ in the third. For $xy=36$, $c=6$, so the general point is $\left(6t,\frac6t\right)$; the point $(4,9)$ is on it since $4\times9=36$ (here $t=\frac23$). It is symmetric about the lines $y=x$ and $y=-x$.

For the parabola, differentiate $y^2=4ax$ implicitly: $2y\frac{dy}{dx}=4a$, so $\frac{dy}{dx}=\frac{2a}{y}=\frac1t.$ For the rectangular hyperbola, $y=\frac{c^2}{x}$ so $\frac{dy}{dx}=-\frac{c^2}{x^2}=-\frac1{t^2}.$ The normal gradient is the negative reciprocal. Using $y-y_1=m(x-x_1)$:

• parabola tangent: $ty=x+at^2$; normal: $y+tx=2at+at^3$
• hyperbola tangent: $x+t^2y=2ct$; normal: $t^3x-ty=c(t^4-1)$. Example: on $y^2=8x$ ($a=2$) at $t=2$, $P=(8,8)$, tangent gradient $\frac12$, normal gradient $-2$, so the normal is $y=-2x+24$, meeting the $x$-axis at $(12,0)$.

Substitute the line into the curve to get a quadratic in $x$. A tangent meets the curve at one point only, so the discriminant is zero. For $y^2=4ax$ and $y=mx+c$: $m^2x^2+(2mc-4a)x+c^2=0,$ and $(2mc-4a)^2-4m^2c^2=0$ simplifies to $c=\frac am.$ Example: $y=mx+2$ and $y^2=8x$ gives $2=\frac2m$, so $m=1$. For $xy=c_0^2$ and $y=mx+k$ the quadratic is $mx^2+kx-c_0^2=0$, with discriminant $k^2+4mc_0^2$, so tangency needs $k^2=-4mc_0^2$ and so $m<0$.

To find a locus, let $P=(x,y)$ be a general point, write the given condition using the distance formula, then simplify. Example: $P$ is equidistant from $F(6,0)$ and the line $x=-2$. Then $(x-6)^2+y^2=(x+2)^2\Rightarrow y^2=16x-32.$ This is a parabola with vertex $(2,0)$, the midpoint of $F$ and the foot of the perpendicular on the directrix. For parametric loci, write $x$ and $y$ in terms of $t$ and eliminate $t$. Example: the tangent at $P\left(4t,\frac4t\right)$ on $xy=16$ is $x+t^2y=8t$; it meets the axes at $(8t,0)$ and $\left(0,\frac8t\right)$, so triangle $OAB$ has area $\frac12(8t)\left(\frac8t\right)=32$, constant.