Ellipse and hyperbolaEdexcel A-Level Further Maths: Revision notes

The ellipse with centre at the origin has Cartesian equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,\quad a>b>0.$ It crosses the axes at $(\pm a,0)$ and $(0,\pm b)$; the semi-major axis is $a$ and the semi-minor axis is $b$. Parametric equations are $x=a\cos t,\quad y=b\sin t,$ which satisfy the Cartesian equation because $\cos^2t+\sin^2t=1$. For $\frac{x^2}{25}+\frac{y^2}{16}=1$, the point with $t=\frac\pi3$ is $\left(\frac52,2\sqrt3\right)$. When $a=b$ the ellipse becomes a circle.

The hyperbola has Cartesian equation $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.$ It has two branches, crossing the $x$-axis at $(\pm a,0)$ and never meeting the $y$-axis. Two parametrisations are on the specification: $x=a\sec t,\ y=b\tan t\quad\text{or}\quad x=a\cosh t,\ y=b\sinh t.$ They work because $\sec^2t-\tan^2t=1$ and $\cosh^2t-\sinh^2t=1$. With $\cosh t\geq1$, the hyperbolic form covers only the branch with $x>0$. For $\frac{x^2}{9}-\frac{y^2}{16}=1$, $t=\frac\pi3$ gives $(3\sec\frac\pi3,4\tan\frac\pi3)=(6,4\sqrt3)$.

Both curves are the locus of a point $P$ with $SP=e\times PM$, where $S$ is a focus, $M$ is the foot of the perpendicular from $P$ to the matching directrix, and $e$ is the eccentricity. For both, the foci are $(\pm ae,0)$ and the directrices are $x=\pm\frac ae$.

• Ellipse: $0<e<1$ and $b^2=a^2(1-e^2)$.
• Hyperbola: $e>1$ and $b^2=a^2(e^2-1)$. Example: $\frac{x^2}{25}+\frac{y^2}{16}=1$ has $16=25(1-e^2)$, so $e=\frac35$, foci $(\pm3,0)$ and directrices $x=\pm\frac{25}{3}$. For $\frac{x^2}{9}-\frac{y^2}{16}=1$, $16=9(e^2-1)$ gives $e=\frac53$, foci $(\pm5,0)$ and directrices $x=\pm\frac95$.

Differentiate implicitly. For the ellipse $\frac{2x}{a^2}+\frac{2y}{b^2}\frac{dy}{dx}=0$, so $\frac{dy}{dx}=-\frac{b^2x}{a^2y}.$ For the hyperbola $\frac{2x}{a^2}-\frac{2y}{b^2}\frac{dy}{dx}=0$, so $\frac{dy}{dx}=\frac{b^2x}{a^2y}.$ The normal gradient is the negative reciprocal. At the point with parameter $t$, the tangents are $\frac{x\cos t}{a}+\frac{y\sin t}{b}=1\ \text{(ellipse)},\qquad \frac{x\sec t}{a}-\frac{y\tan t}{b}=1\ \text{(hyperbola)}.$ Example: on $\frac{x^2}{4}-\frac{y^2}{9}=1$ at $(4,3\sqrt3)$, $\frac{dy}{dx}=\frac{9(4)}{4(3\sqrt3)}=\sqrt3$, so the tangent is $y=\sqrt3(x-1)$, which meets the $x$-axis at $(1,0)$.

Substitute $y=mx+c$ into the curve and set the discriminant of the quadratic in $x$ to zero. This gives $c^2=a^2m^2+b^2\ \text{(ellipse)},\qquad c^2=a^2m^2-b^2\ \text{(hyperbola)}.$ Example: for $\frac{x^2}{4}-\frac{y^2}{9}=1$ and gradient $2$, $c^2=4(4)-9=7$, so $c=\pm\sqrt7$, giving the two tangents $y=2x\pm\sqrt7$. For the hyperbola, if $a^2m^2<b^2$ then $c^2$ would be negative, so no tangent of that gradient exists.

For a locus, let $P=(x,y)$, write the condition with distances, then square and simplify. If $SP=e\times PM$ the locus is an ellipse for $e<1$, a hyperbola for $e>1$ and a parabola for $e=1$. Example: $P$ is such that its distance from $S(3,0)$ is $\frac32$ of its distance from the line $x=\frac43$. $(x-3)^2+y^2=\frac94\left(x-\frac43\right)^2\Rightarrow y^2=\frac54x^2-5\Rightarrow\frac{x^2}{4}-\frac{y^2}{5}=1.$ This is a hyperbola with $a=2$, and $e=\frac32$ is consistent with $b^2=a^2(e^2-1)=5$. For parametric loci, write $x$ and $y$ in terms of $t$ and eliminate $t$ using $\cos^2t+\sin^2t=1$ or $\sec^2t-\tan^2t=1$.